Post on 31-Mar-2021
SOURCE IDENTIFICATION PROBLEMS FOR
HYPERBOLIC DIFFERENTIAL AND DIFFERENCE
EQUATIONS
A THESIS SUBMITTED TO THE GRADUATE
SCHOOL OF APPLIED SCIENCES
OF
NEAR EAST UNIVERSITY
By
FATHI S. A. EMHARAB
In Partial Fulfillment of the Requirements for
the Degree of Doctor of Philosophy
in
Mathematics
NICOSIA, 2019
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019
SOURCE IDENTIFICATION PROBLEMS FOR
HYPERBOLIC DIFFERENTIAL AND DIFFERENCE
EQUATIONS
A THESIS SUBMITTED TO THE GRADUATE
SCHOOL OF APPLIED SCIENCES
OF
NEAR EAST UNIVERSITY
By
FATHI S. A. EMHARAB
In Partial Fulfillment of the Requirements for
the Degree of Doctor of Philosophy
in
Mathematics
NICOSIA, 2019
Fathi S. A. Emharab: SOURCE IDENTIFICATION PROBLEMS FOR HYPERBOLIC
DIFFERENTIAL AND DIFFERENCE EQUATIONS
Approval of Director of Graduate School of
Applied Sciences
Prof. Dr. Nadire ÇAVUŞ
We certify this thesis is satisfactory for the award of the degree of Doctor of Philosophy of
Science in Mathematics
Examining Committee in Charge:
Prof. Dr. Evren Hinçal Committee Chairman, Department of
Mathematics, NEU
Prof. Dr. Allaberen Ashyralyev Supervisor, Department of Mathematics, NEU
Assoc. Prof. Dr. Deniz Ağırseven Department of Mathematics, Trakya University
Assoc. Prof. Dr. Okan Gerçek Department of Computer Engineering, Girne
American University
Assist. Prof. Dr. Bilgen Kaymakamzade Department of Mathematics, NEU
I hereby declare that all information in this document has been obtained and presented in
accordance with academic rules and ethical conduct. I also declare that, as required by these
rules and conduct, I have fully cited and referenced all material and results that are not
original to this work.
Name, Last name: Fathi S. A. Emharab
Signature:
Date:
iii
ACKNOWLEDGEMENT
The completion of this thesis would not be achieved without the help, backing, support,
and advice of some important individuals. There are no words that can express how
thankful and grateful I am to them.
Foremost, I would like to express my sincere gratitude to my advisor Prof. Dr. Allaberen
Ashyralyev for the continuous support of my Ph.D study and research, for his patience,
motivation, enthusiasm, and immense knowledge. His guidance helped me in all the time
of research and writing of this thesis. I could not have imagined having a better advisor and
mentor for my Ph.D study.
Besides my advisor, I would like to thank some of the great Mathematicians of our time in
persons of Prof. Dr. Charyyar Ashyralyyev for his helpful discussions and his guidance in
Matlab Implementation, Prof. Dr. Evren Hincal and Prof. Dr. Adiguzel Dosiyev for their
guidance, encouragement, and insightful comments. My appreciation also goes to staffs of
Mathematics Department Near East University, especially Dr. Hediye Sarikaya, Dr. Bilgen
Kaymakamzade, and Dr. Nuriye Sancar. Also I thank my friends Kheireddine Belakroum,
Farouk Saad, and Ayman Hamad. I thank you all for your support.
I am extremely thankful for the funding provided by the Ministry of Higher Education and
Scientific Research of Lybia.
Last but not the least, I would like to thank my family: my parents and to my wife and
brothers and sisters for supporting me spiritually throughout writing this thesis and my life
in general.
v
ABSTRACT
In the present study, a source identification problem with local and nonlocal conditions for
a one-dimensional hyperbolic equation is investigated. Stability estimates for the solutions
of the source identification problems are established. Furthermore, a first and second order
of accuracy difference schemes for the numerical solutions of the source identification
problems for hyperbolic equations with local and nonlocal conditions are presented.
Stability estimates for the solutions of difference schemes are established. Then, these
difference schemes are tested on examples and some numerical results are presented.
Keywords: Source identification problem; hyperbolic differential equations; difference
schemes; local and nonlocal conditions; stability; accuracy
vi
ÖZET
Bu tezde bir boyutlu bir hiperbolik denklem için yerel ve yerel olmayan koşullu bir kaynak
tanımlama problemi araştırılmıştır. Kaynak tanımlama probleminin çözümü için kararlılık
kestirimleri oluşturulmuştur. Ayrıca, hiperbolik denklemler için yerel ve yerel olmayan
koşullu kaynak tanımlama problemlerinin sayısal çözümleri için birinci ve ikinci dereceden
doğruluklu fark şemaları sunulmuştur. Fark şemalarının çözümleri için kararlılık
kestirimleri oluşturulmuştur. Daha sonra, bu fark şemaları örnekler üzerinde test edilip bazı
sayısal sonuçlar verilmiştir.
Anahtar Kelimeler: Kaynak tanımlama problemi; hiperbolik diferansiyel denklemler; fark
şemaları; yerel ve yerel olmayan koşullar; kararlılık; doğruluk
vii
TABLE OF CONTENTS
ACKNOWLEDGEMENT .............................................................................................. iii
ABSTRACT ..................................................................................................................... v
ÖZET ……………………………………………………………………….….……..….. vi
TABLE OF CONTENTS ………………………………………….………………..….. vii
LIST OF TABLES ……………………………………………………………..……….. ix
CHAPTER 1: INTRODUCTION
1.1 History …………………………………………………………………….……..…… 1
1.2 Methods of Solution of Identification Problem ……………………………….……… 5
1.3 The Aim of Thesis …………………………………………………………………… 22
CHAPTER 2: STABILITY OF THE HYPERBOLIC DIFFERENTIAL AND
DIFFERENCE EQUATION WITH LOCAL CONDITIONS
2.1 Introduction ………………………………………………………………….…..….. 23
2.2 Stability of the Differential Problem (1) …………………………………….…....… 23
2.3 Stability of the Difference Scheme ……………………………………….……....… 33
2.3.1 The first order of accuracy difference scheme ….……….…………….…….... 37
2.3.2 The second order of accuracy difference scheme ……….…………….…..….. 45
2.4 Numerical Experiments ……………………………..…………………….….…….. 55
CHAPTER 3: STABILITY OF THE HYPERBOLIC DIFFERENTIAL AND
DIFFERENCE EQUATION WITH NONLOCAL CONDITIONS
3.1 Introduction………………………………………………….……………………… 65
3.2 Stability of the Differential Problem ……………………………….…..…..….….... 65
3.3 Stability of the Difference Scheme …….……………………………….......…..….. 68
3.3.1 The first order of accuracy difference scheme ……………………........…..... 70
viii
3.3.2 The second order of accuracy difference scheme ……….……….…………. 73
3.4 Numerical Experiments ………………………………………….…………...……. 79
CHAPTER 4: CONCLUSION
Conclusion ………………………………..………………...………………….…...…... 89
REFERENCES………………………………………………………...…………….... 90
APPENDICES
Appendix 1: …………………………..……………………………………………….... 95
Appendix 2: …………………..……………….……………………………..….……... 100
Appendix 3: …………………………..……………………………..………….….…... 107
Appendix 4: …………………..……………….…………………………….….….…... 114
Appendix 5: …………………………..………………………………….….…..……... 118
ix
LIST OF TABLES
Table 1: Error analysis for difference scheme (2.69)…………………………………... 58
Table 2: Error analysis for difference scheme (2.75)…………………………………... 64
Table 3: Error analysis for difference scheme (2.76)…………………………………... 64
Table 4: Error analysis for difference scheme (3.42)……………………….……….… 82
Table 5: Error analysis for difference scheme (3.48)……………………………….…. 88
CHAPTER 1
INTRODUCTION
1.1 History
The studies of well-posed and ill-posed boundary value problems for hyperbolic and telegraph
partial differential equations are driven not only by a theoretical interest but also by the fact
of several phenomena in engineering and various fields of physics and applied sciences. In
mathematical modelling, hyperbolic and telegraph partial differential equations are used
together with boundary conditions specifying the solution on the boundary of the domain. In
some cases, classical boundary conditions cannot describe a process or phenomenon precisely.
Therefore, mathematical models of various physical, chemical, biological, or environmental
processes often involve nonclassical conditions. Such conditions are usually identified as
nonlocal boundary conditions and reflect situations when the data on the boundary of domain
cannot be measured directly, or when the data on the inside of the domain. Of great interest is
the study of absolutely stable difference schemes of a high order of accuracy for hyperbolic
partial differential equations, in which stability was established without any assumptions with
respect to the grid steps and such type of stability inequalities for the solutions of the first
order of accuracy difference scheme for the differential equations of hyperbolic type were
established for the first time ( Sobolevskii and Chebotareva, 1977).
The survey paper contains the results on the local and nonlocal well-posed problems for second
order differential and difference equations. Results on the stability of differential problems for
hyperbolic equations and of difference schemes for approximate solution of the hyperbolic
problems were presented (Ashyralyev et al., 2015).
Identification problems take an important place in applied sciences and engineering, and have
been studied by many authors (Belov, 2002; Gryazin et al., 1999; Isakov, 1998; Kabanikhin
and Krivorotko, 2015; Prilepko, Orlovsky and Vasin, 2000). The theory and applications of
source identification problems for partial differential equations have been given in various
papers ( Anikonov, 1996; Ashyralyev and Ashyralyyev, 2014; Ashyralyyev, 2014; Ashyralyyev
and Demirdag, 2012; Kozhanov, 1997; Orlovskii, 2008; Orlovskii and Piskarev, 2013).
1
In particular, Kozhanov, (1997) applied a new approach for solving elliptic equations which is
based on the transition to equations of composite type. The obtained results on solvability of
linear inverse problems for elliptic equations are based on the solvability and the properties of
solutions of boundary value problems for equations of composite type. The inverse problem
of finding the source in an abstract second-order elliptic equation on a finite interval was
studied by (Orlovskii, 2008). The additional information given is the value of the solution
at an interior point of the interval. Moreover, existence, uniqueness, and Fredholm property
theorems for the inverse problem were proved. The authors investigated an inverse problem for
an elliptic equation in a Banach space with the Bitsadze-Samarskii conditions. The suggested
approach uses the notion of a general approximation scheme, the theory of C0-semigroups of
operators and methods of functional analysis (Orlovskii and Piskarev, 2013).
The well-posedness of the unknown source identification problem for a parabolic equation
has been well investigated when the unknown function p is dependent on the space variable
(Ashyralyev, 2011; Ashyralyev, Erdogan and Demirdag, 2012; Choulli and Yamamoto, 1999;
Èidel’man, 1978; Kostin, 2013). Nevertheless, when the unknown function p is dependent
on t, the well-posedness of the source identification problem for a parabolic equation has
been investigated by (Ashyralyev and Erdogan, 2014; Borukhov and Vabishchevich, 2000;
Dehghan, 2003; Erdogan and Sazaklioglu, 2014; Ivanchov, 1995; Saitoh, Tuan and Yamamoto,
2003; Samarskii and Vabishchevich, 2008). Moreover, the well-posedness of the source
identification problem for a delay parabolic equation has also been given by (Ashyralyev and
Agirseven, 2014; Blasio and Lorenzi, 2007). The authors studied the inverse problems in
determining the coefficients of the equation for the kinetic Boltzmann equation. The Cauchy
problem and the boundary value problem for states close to equilibrium have been considered.
Theorems of the existence and uniqueness of the inverse problems were proved (Orlovskii and
Prilepko, 1987).
The solvability of the inverse problems in various formulations with various overdetermination
conditions for telegraph and hyperbolic equations were studied in many works (Anikonov,
1976; Ashyralyev and Çekiç, 2015; Kozhanov and Safiullova, 2010; Kozhanov and Safiullova,
2017; Kozhanov and Telesheva, 2017). In particular, the well-posedness of the source
2
identification problem for a telegraph equation with unknown parameter pd2v(t)
dt2 + αdv(t)
dt + Av (t) = p + f (t) ,0 ≤ t ≤ T,
v (0) = ϕ, v′ (0) = ψ, v (T) = ζ
in aHilbert space H with the self-adjoint positive-definite operator Awas proved by (Ashyralyev
and Çekiç, 2015). Here ϕ,ψ and ζ are given elements of H. They established stability estimates
for the solution of this problem. In applications, three source identification problems for
telegraph equations are developed. The authors studied the solvability of the inverse problems
on finding a solution v (x, t) and an unknown coefficient c for a telegraph equation
vtt − ∆v + cv = f (x, t) .
Theorems on the existence of the regular solutions are proved. The feature of the problems is a
presence of new overdetermination conditions for the considered class of equations (Kozhanov
and Safiullova, 2017). The authors studied solvability of the parabolic and hyperbolic inverse
problems of finding a solution together with an unknown right-hand side when the general
overdetermination condition is given. Some theorems of unique existence of regular solutions
were proved ( Kozhanov and Safiullova, 2010). The authors considered nonlinear inverse
coefficient problems for nonstationary higher-order differential equations of pseudohyperbolic
type (Kozhanov and Telesheva, 2017). More precisely, they study the problems of determining
both the solution of the corresponding equation, an unknown coefficient at the solution or at
the time derivative of the solution in the equation. A distinctive feature of these problems is
the fact that the unknown coefficient is a function of time only. Integral overdetermination is
used as an additional condition. The existence theorems of regular solutions (those solutions
that have all generalized derivatives in the sense of S. L. Sobolev) were proved. The technique
of the proof relies on the transition from the original inverse problem to a new direct problem
for an auxiliary integral-differential equation, and then on the proof of solvability of the latter
and construction of some solution of the original inverse problem from a solution of the
auxiliary problem. A theorem on the uniqueness for solutions to an inverse problem for the
wave equation has been proved (Anikonov, 1976). Finally, some new representations were
given for the solutions and coefficients of the equations of mathematical physics (Anikonov,
3
1995; Anikonov, 1996; Anikonov and Neshchadim, 2011; Anikonov and Neshchadim, 2012
a, 2012b). Their main direction of study was to search for reciprocal formulas connecting
solutions and coefficients, and involving arbitrary functions as well as functions satisfying
some differential relations. They gave such formulas for evolution equations of first and
second order in time, in particular for parabolic and hyperbolic equations in the linear and
nonlinear cases.
In this thesis, we consider the time-dependent source identification problem for a one-
dimensional hyperbolic equation with local conditions
∂2u (t, x)∂t2 −
∂
∂x
(a (x)
∂u (t, x)∂x
)= p (t) q (x) + f (t, x) ,
x ∈ (0, l) , t ∈ (0,T) ,
u (0, x) = ϕ (x) ,ut (0, x) = ψ (x) , x ∈ [0, l] ,
u (t,0) = u (t, l) = 0,∫ l0 u (t, x) dx = ζ (t) , t ∈ [0,T]
(1.1)
and with nonlocal conditions
∂2u (t, x)∂t2 −
∂
∂x
(a (x)
∂u (t, x)∂x
)+ δu (t, x) = p (t) q (x) + f (t, x) ,
x ∈ (0, l) , t ∈ (0,T) ,
u (0, x) = ϕ (x) ,ut (0, x) = ψ (x) , x ∈ [0, l] ,
u (t,0) = u (t, l) ,ux (t,0) = ux (t, l) ,∫ l0 u (t, x) dx = ζ (t) , t ∈ [0,T] ,
(1.2)
where u (t, x) and p (t) are unknown functions, a (x) ≥ a > 0, δ ≥ 0, f (t, x) , ζ (t) , ϕ (x) and
ψ (x) are sufficiently smooth functions, and q (x) is a sufficiently smooth function assuming∫ l0 q (x) dx , 0, and q (0) = q (l) = 0 for (1.1), q (0) = q (l) , q′ (0) = q′ (l) for (1.2). At the
same time, we note that the inverse problems (1.1) and (1.2) for the hyperbolic equation were
not studied before. Basic results of this thesis have been published by the following papers
(Ashyralyev and Emharab, 2017; Ashyralyev and Emharab, 2018a, 2018b, 2018c, 2018d).
Some results of this work were presented in Mini-symposium "Inverse Ill-posed Problems
and its applications" of VI congress of Turkic World Mathematical Society (TWMS 2017),
and 2nd International Conference of Mathematical Sciences, 2018.
4
1.2 Methods of Solution of Source Identification Problem
It is known that local and nonlocal boundary value problems for second order partial differential
equations can be solved analytically by Fourier series, Laplace transform and Fourier transform
methods. Now, let us illustrate these three different analytical methods by examples. Let us
see how to apply the classic methods, namely, Fourier series, Fourier transform and Laplace
transform for obtaining the solution of source identification problem for hyperbolic equations
on some examples.
Example 1.2.1 Obtain the Fourier series solution of the following source identification
problem
∂2u (t, x)∂t2 −
∂2u (t, x)∂x2 = p (t) sin x + e−t sin x,
0 < x < π,0 < t < 1,
u (0, x) = sin x,ut (0, x) = − sin x,0 ≤ x ≤ π,
u (t, π) = u (t,0) = 0,∫ π
0 u (t, x) dx = 2e−t,0 ≤ t ≤ 1,
(1.3)
for a one dimensional hyperbolic equation.
Solution. In order to solve the problem, we consider the Sturm-Liouville problem
uxx − λu (x) = 0,0 < x < π, u (0) = u (x) = 0,
generated by the space operator of problem (1.3). It is easy to see that the solution of this
Sturm-Liouville problem is
λk = −k2,uk(x) = sin k x, k = 1,2, ....
Therefore, we will seek solution u(t, x) using by the Fourier series
u (t, x) =∞∑
k=1Ak (t) sin k x. (1.4)
Here Ak (t) , k = 1,2... are unknown functions. Putting (1.4) into the equation (1.3) and using
given initial and boundary conditions, we obtain
∞∑k=1
A′′k (t) sin k x +∞∑
k=1k2 Ak (t) sin k x = p (t) sin x + e−t sin x,
5
u (0, x) =∞∑
k=0Ak (0) sin k x = sin x,
ut (0, x) =∞∑
k=0A′k (0) sin k x = − sin x,
∫ π
0u (t, x) dx =
∫ π
0
∞∑k=1
Ak (t) sin k xdx = −∞∑
k=1
Ak (t) cos k xk
π0
=
∞∑k=1
Ak (t)
[1 + (−1)k+1
k
]= 2e−t .
Equating the coefficients of sin k x, k = 1,2... to zero, we get
A′′k (t) + k2 Ak (t) = 0, k , 1, A′′1 (t) + A1 (t) = p (t) + e−t,0 < t < 1,
Ak (0) = 0, A′k (0) = 0, k , 1, A1 (0) = 1, A′1 (0) = −1,∞∑
k=1Ak (t)
[1 + (−1)k+1
k
]= 2e−t,0 < t < 1. (1.5)
First, we will obtain Ak (t) for k , 1. It is easy to see that Ak (t) is the solution of the
following Cauchy problem
A′′k (t) + k2 Ak (t) = 0, 0 < t < 1, Ak (0) = 0, A′k (0) = 0,
for the second order differential equations. Its solution is Ak (t) ≡ 0, k , 1. From that and
formula (1.5), we get
−2A1 (t) = −2e−t .
Therefore,
A1 (t) = e−t . (1.6)
Second,we will obtain p (t). It is clear that A1 (t) is solution of the following Cauchy problem
A′′1 (t) + A1 (t) = p (t) + 2e−t, 0 < t < 1, A1 (0) = 1, A′1 (0) = −1, (1.7)
for the second order differential equations. Applying (1.6) and (1.7),we obtain
p (t) = e−t .
6
Then, (1.4) becomes u (t, x) = A1 (t) sin x = e−t sin x. Therefore, the exact solution of problem
(1.3) is (u (t, x) , p (t)) =(e−t sin x, e−t ) .
Note that using similar procedure one can obtain the solution of the following source
identification problem
∂2u(t, x)∂t2 −
n∑r=1
αr∂2u(t, x)∂x2
r= p(t)q(x) + f (t, x),
x = (x1, ..., xn) ∈ Ω, 0 < t < T,
u(0, x) = ϕ(x), ut(0, x) = ψ (x) , x ∈ Ω,
u(t, x) = 0, 0 ≤ t ≤ T, x ∈ S,∫· · ·
∫x∈Ω
u (t, x) dx1...dxn = ξ(t),0 ≤ t ≤ T,
(1.8)
for the multidimensional hyperbolic partial differential equation. Assume that αr > α > 0
and f (t, x) ,q (x) , (t ∈ (0,T) , x ∈ Ω), ϕ(x),ψ (x) ,(x ∈ Ω
), ξ(t), (t ∈ [0,T]) are given smooth
functions. Here and in future Ω is the unit open cube in the n−dimensional Euclidean space
Rn (0 < xk < 1,1 ≤ k ≤ n) with the boundary S and Ω = Ω ∪ S.
Example 1.2.2 Obtain the Fourier series solution of the following source identification
problem
∂2u (t, x)∂t2 +
∂u (t, x)∂x2 = p (t) (1 + sin 2x) + 4e−t sin 2x,
0 < t < 1,0 < x < π,
u (0, x) = 1 + sin 2x,ut (0, x) = − (1 + sin 2x) , 0 ≤ x ≤ π,
u (t,0) = u (t, π) ,ux (t,0) = ux (t, π) , 0 ≤ t ≤ 1,∫ π
0 u (t, x) dx = e−tπ, 0 ≤ t ≤ 1,
(1.9)
for a one dimensional hyperbolic equation.
Solution. In order to solve this problem, we consider the Sturm-Liouville problem
uxx − λu (x) = 0,0 < x < π, u (0) = u (x) , ux (0) = ux (π)
7
generated by the space operator of problem (1.9). It is easy to see that the solution of this
Sturm-Liouville problem is
λk = 4k2,uk (x) = cos 2k x, k = 0,1,2, ..., uk (x) = sin 2k x, k = 1,2, ....
Therefore, we will seek solution u(t, x) using by the Fourier series
u (t, x) =∞∑
k=0Ak (t) cos 2k x +
∞∑k=1
Bk (t) sin 2k x, (1.10)
where Ak (t) , k = 0,1,2, ... and Bk (t) , k = 1,2, ... are unknown functions, Putting (1.10) into
(1.9) and using given initial and boundary conditions, we get
∞∑k=0
A′′k (t) cos 2k x +∞∑
k=1B′′k (t) sin 2k x +
∞∑k=0
4k2 Ak (t) cos k x
+
∞∑k=1
4k2Bk (t) sin 2k x = p (t) (sin 2x + 1) + 4e−t sin 2x,
u (0, x) =∞∑
k=0Ak (0) cos 2k x +
∞∑k=1
Bk (0) sin 2k x = sin 2x + 1,
ut (0, x) =∞∑
k=0A′k (0) cos 2k x +
∞∑k=1
B′k (0) sin 2k x = − sin 2x − 1,
∫ π
0u (t, x) dx =
∫ π
0
[∞∑
k=0Ak (t) cos 2k x +
∞∑k=1
Bk (t) sin 2k x
]dx
= A0 (t) π +∞∑
k=1
Ak (t) sin 2k x2k
]π0
−
∞∑k=1
Bk (t) cos 2k x2k
]π0
= A0 (t) π = e−tπ.
From that it follows that A0 (t) = e−t . Equating the coefficients of cos k x, k = 0,1,2, ... and
sin k x, k = 1,2, ... to zero, we getA′′k (t) + 4k2 Ak (t) = 0, 0 < t < 1,
Ak (0) = 0, A′k (0) = 0, k , 0,(1.11)
A′′0 (t) = p (t) , 0 < t < 1,
A0 (0) = 1, A′0 (0) = −1,(1.12)
8
B′′k (t) + 4k2Bk (t) = 0, 0 < t < 1,
Bk (0) = 0,B′k (0) = 0,(1.13)
B′′1 (t) + 4B1 (t) = p (t) + 4e−t, 0 < t < 1,
B1 (0) = 1,B′1 (0) = −1.(1.14)
First, we obtain p(t). Applying problem (1.12) and A0(t) = e−t,we get
p (t) = e−t .
Second, we obtain Ak(t), k , 0. It is clear that for k , 0, Ak(t) is the solution of the initial
value problems (1.11) and (1.12). The auxiliary equation is
m2 + 4k2 = 0.
We have two roots
m1 = 2ik,m2 = −2ik
Therefore,
Ak (t) = c1 cos 2kt + c2 sin 2kt.
Applying initial conditions Ak (0) = A′k (0) = 0,we get
Ak (0) = c1 = 0,
A′k (0) = 2kc2 = 0.
Then c1 = c2 = 0 and Ak (t) = 0, k , 0. Third, we obtain Bk(t). It is clear that for k , 1,Bk(t)
is the solution of the initial value problem (1.13). The auxiliary equation is
m2 + 4k2 = 0.
We have two roots
m1 = 2ik,m2 = −2ik .
Therefore,
Bk (t) = c1 cos 2kt + c2 sin 2kt.
9
Applying initial conditions Bk (0) = B′k (0) = 0,we get
Bk (0) = c1 = 0,
B′k (0) = 2kc2 = 0.
Then c1 = c2 = 0 and Bk (t) = 0. Now, we obtain B1 (t) from (1.14), and p (t) = e−t . It is
the Cauchy problem for the second order linear differential equation. We will seek B1 (t) by
formula
B1 (t) = BC (t) + BP (t) ,
where BC (t) is general solution of the homogeneous differential equation B′′1 (t)+ 4B1 (t) = 0
and BP (t) is particular solution of non-homogeneous differential equation. The auxiliary
equation is
m2 + 4 = 0.
We have two roots
m1 = 2ik,m2 = −2ik .
Therefore,
Bc (t) = c1 cos 2kt + c2 sin 2kt.
Since ±i2 , −1,we put
Bp (t) = e−ta.
Therefore,
ae−t + 4ae−t = 5e−t .
From that it follows a = 1 and
Bp (t) = e−t .
Thus,
B1 (t) = c1 cos 2kt + c2 sin 2kt + e−t .
Applying initial conditions B1 (0) = 1,B′1 (0) = −1,we get
B1 (0) = c1 + 1 = 1, B′1 (0) = 2kc2 − 1 = −1.
10
From that it follows
B1 (t) = e−t .
Therefore,
u (t, x) = e−t (sin 2x + 1) .
So, the exact solution of problem (1.9) is
(u (t, x) , p (t)) =(e−t (sin 2x + 1) , e−t ) .
Note that using similar procedure one can obtain the solution of the following source
identification problem
∂2u(t, x)∂t2 −
n∑r=1
αr∂2u(t, x)∂x2
r= p (t) q (x) + f (t, x),
x = (x1, ..., xn) ∈ Ω, 0 < t < T,
u(0, x) = ϕ(x),ut(0, x) = ψ (x) , x ∈ Ω,
u(t, x)|S1 = u(t, x)|S2 ,∂u(t, x)∂m
S1
=∂u(t, x)∂m
S2
,∫· · ·
∫x∈Ω
u (t, x) dx1...dxn = ξ(t),0 ≤ t ≤ T,
(1.15)
for the multidimensional hyperbolic partial differential equation. Assume that αr > α > 0 and
f (t, x) ,q (x) , (t ∈ (0,T) , x ∈ Ω) ,ψ (x) , ξ(t),(t ∈ [0,T] , x ∈ Ω
)are given smooth functions.
Here S = S1 ∪ S2,S1 ∩ S2 = ∅, and m is the normal vector to S1 and S2.
However Fourier series method described in solving (1.8) and (1.15) can be used only in the
case when (1.8) and (1.15) have constant coefficients.
Second, we consider the Laplace transform method for solution of the source identification
problem for hyperbolic differential equation.
Example 1.2.3 Obtain the Laplace transform solution of the following source identifica-
11
tion problem
∂2u (t, x)∂t2 −
∂2u (t, x)∂x2 + 2u (t, x) = p (t) e−x + e−t−x,
0 < x < ∞,0 < t < 1,
u (0, x) = e−x,ut (0, x) = −e−x,0 ≤ x < ∞,
u (t,0) = e−t,ux (t,0) = −e−t,0 ≤ t ≤ 1,∫ ∞0 u (t, x) dx = e−t,0 ≤ t ≤ 1,0 ≤ x < ∞,
(1.16)
for a one dimensional hyperbolic equation.
Solution. We will denote
L u(t, x) = u(t, s).
Using formula
L e−x =1
s + 1(1.17)
and taking the Laplace transform of both sides of the differential equation and conditions
u(t,0) = e−t,ux(t,0) = −e−t , we can write
L utt (t, x) − L uxx (t, x) + 2L u (t, x) =[p (t) + e−t ] L e−x ,
L u (0, x) = L e−x ,L ut (0, x) = −L e−x
or utt (t, s) − s2u (t, s) + su (t,0) + ux (t,0) + 2u (t, s) = p (t) 1
s+1 + e−t 1s+1,
u(0, s) =1
s + 1,ut(0, s) = −
1s + 1
.
Therefore, we get the following problemutt (t, s) +
(2 − s2) u (t, s) + se−t − e−t = p (t) 1
s+1 + e−t 1s+1,
u(0, s) =1
s + 1,ut(0, s) = −
1s + 1
.
Applying the condition ∫ ∞
0u (t, x) dx = e−t,0 ≤ t ≤ 1,
and the definition of the Laplace transform, we get
u (t,0) = e−t,0 ≤ t ≤ 1, (1.18)
12
It is clear that u(t, s) is solution of the following source identification problemutt (t, s) +
(2 − s2) u (t, s) = p (t) 1
s+1 +2−s2
s+1 e−t,
u (0, s) = 1s+1,ut (0, s) = 1
s+1 .(1.19)
Applying the D’Alembert’s formula, we obtain
u (t, s) =1
s + 1cos
√2 − s2t −
1s + 1
1√
2 − s2sin
√2 − s2t (1.20)
+1
√2 − s2
∫ t
0sin
√2 − s2 (t − y)
p (y)
1s + 1
+2 − s2
s + 1e−y
dy.
Now, we will apply the condition (1.18) with (1.20), we get
e−t = cos√
2t −1√
2sin√
2t +1√
2
∫ t
0sin√
2 (t − y) p (y) + 2e−y dy. (1.21)
Taking the first and second order derivatives, we get
−e−t = −√
2 sin√
2t − cos√
2t +∫ t
0cos√
2 (t − y) p (y) + 2e−y dy,
e−t = −2 cos√
2t +√
2 sin√
2t
−√
2∫ t
0sin√
2 (t − y) p (y) + 2e−y dy + p (t) + 2e−t . (1.22)
Applying (1.21) and (1.22), we get
e−t = −2 cos√
2t +√
2 sin√
2t + p (t) + 2e−t
−√
2√
2e−t − cos
√2t +
1√
2sin√
2t.
From that it follows
p (t) = e−t . (1.23)
Finally, applying (1.20) and (1.23), we get
u (t, s) =1
s + 1cos
√2 − s2t −
1s + 1
1√
2 − s2sin
√2 − s2t
+1
√2 − s2
∫ t
0sin
√2 − s2 (t − y)
e−y
(3 − s2
s + 1
)dy.
13
Applying the formula∫ t
0sin
√2 − s2 (t − y) e−ydy =
sin√
2 − s2t + e−t√
2 − s2 −√
2 − s2 cos√
2 − s2t3 − s2 ,
we get
u (t, s) =1
s + 1cos
√2 − s2t −
1s + 1
1√
2 − s2sin
√2 − s2t
+1
√2 − s2
3 − s2
s + 1
[sin√
2 − s2t + e−t√
2 − s2 −√
2 − s2 cos√
2 − s2t3 − s2
]=
1s + 1
cos√
2 − s2t −1
s + 11
√2 − s2
sin√
2 − s2t
+1
√2 − s2
1s + 1
sin√
2 − s2t +e−t
s + 1−
1s + 1
cos√
2 − s2t = e−t 1s + 1
. (1.24)
From that it follows that
u (t, x) = e−tL−1
1s + 1
= e−t−x .
Therefore, the exact solution of problem (1.16) is
(u (t, x) , p (t)) =(e−t−x, e−t ) .
Example 1.2.4 Obtain the Laplace transform solution of the following source identification
problem
∂2u (t, x)∂t2 +
∂u (t, x)∂t
= p (t) (1 + sin 2x) + 4e−t sin 2x,
t > 0,0 < x < π,
u (0, x) = 1 + sin 2x, ut (0, x) = − (1 + sin 2x) , 0 ≤ x ≤ π,
u (t,0) = u (t, π) ,ux (t,0) = ux (t, π) , t ≥ 0,∫ π
0 u (t, x) dx = e−tπ, 0 ≤ t ≤ 1,
(1.25)
for a one dimensional hyperbolic equation.
Solution. We will denote
L u(t, x) = u(s, x).
14
Using formula
Le−t = 1
s + 1and taking the Laplace transform of both sides of the differential equation and conditions
u(0, x) = 1 + sin 2x,ut(0, x) = −(1 + sin 2x), we can writes2u (s, x) − u (s,0) − ux (t,0) + uxx (s, x)
= p (s) (1 + sin 2x) + 41+s sin 2x,
u (s,0) = u (s, π) ,ux (s,0) = ux (s, π) , 0 ≤ t ≤ 1.
Therefore, we get the following problem
s2u (s, x) − s (1 + sin 2x) + (1 + sin 2x) − uxx (s, x)
= p (s) (1 + sin 2x) + 41+s sin 2x,
u (t,0) = u (t, π) ,ux (t,0) = ux (t, π) ,∫ π
0 u (s, x) dx = π1+s , 0 ≤ t ≤ 1.
(1.26)
In order to solve this problem, we consider the Sturm-Liouville problem
uxx − λu (x) = 0,0 < x < π, u (0) = u (x) , ux (0) = ux (π)
generated by the space operator of problem (1.26). It is easy to see that the solution of this
Sturm-Liouville problem is
λk = 4k2,uk (x) = cos 2k x, k = 0,1,2, ..., uk (x) = sin 2k x, k = 1,2, ....
Then, we will obtain the Fourier series solution of problem (1.26) by formula
u (s, x) =∞∑
k=0Ak (s) cos 2k x +
∞∑k=1
Bk (s) sin 2k x, (1.27)
where Ak (t) , k = 0,1,2, ... and Bk (t) , k = 1,2, ...are unknown functions, Putting (1.27) into
(1.26) and using given initial and boundary conditions, we get
s2∞∑
k=0Ak (s) cos 2k x − s (1 + sin 2x) + (1 + sin 2x)
+s2∞∑
k=1Bk (s) sin 2k x +
∞∑k=0
4k2 Ak (s) cos k x
15
+
∞∑k=1
4k2Bk (s) sin 2k x = p (s) (1 + sin 2x) +4
1 + ssin 2x,
u (0, x) =∞∑
k=0Ak (0) cos 2k x +
∞∑k=1
Bk (0) sin 2k x = 1 + sin 2x,
ut (0, x) =∞∑
k=0A′k (0) cos 2k x +
∞∑k=1
B′k (0) sin 2k x = −1 − sin 2x,
∫ π
0u (s, x) dx =
∫ π
0
[∞∑
k=0Ak (s) cos 2k x +
∞∑k=1
Bk (s) sin 2k x
]dx
= A0 (s) π +∞∑
k=1
Ak (s) sin 2k x2k
]π0
−
∞∑k=1
Bk (s) cos 2k x2k
]π0
= A0 (s) π =π
1 + s.
From that it follows A0 (s) = 11+s . Equating the coefficients of cos k x, k = 0,1,2, ... and
sin k x, k = 1,2, ... to zero, we gets2 Ak (s) + 4k2 Ak (s) = 0, k , 0,
Ak (0) = 0, A′k (0) = 0,and Ak (s) = 0, k , 0,(1.28)
s2 A0 (s) − s + 1 = p (s) ,
A0 (0) = 1, A′0 (0) = −1,(1.29)
s2Bk (s) + 4k2Bk (s) = 0,
Bk (0) = 0,B′k (0) = 0,and Bk (s) = 0, k , 0,(1.30)
s2B1 (s) + 4B1 (s) − s + 1 = p (s) + 4
s+1,
B1 (0) = 1, B′1 (0) = 1.(1.31)
First, we obtain p(s). Applying problem (1.29) and A0(t) = 11+s ,we get
p (s) =1
1 + s.
Second, we obtain B1 (t) from (1.31) and p (s) = 11+s ,where(
s2 + 4)
B1 (s) − s + 1 =1
1 + s+
4s + 1
,
Thus,
B1 (s) =1
1 + s.
16
Then, (1.27) becomes
u (s, x) =1
1 + s+
11 + s
sin 2x,
From that it follows
u (t, x) = L−1
11 + s
+1
1 + ssin 2x
= e−t (1 + sin 2x) .
Therefore, the exact solution of problem (1.25) is
u (t, x) , p (t) =e−t (1 + sin 2x) , e−t .
Note that using similar procedure one can obtain the solution of the following source
identification problem
∂2u(t, x)∂t2 −
n∑r=1
ar∂2u(t, x)∂x2
r= p(t)q(x) + f (t, x),
x = (x1, ..., xn) ∈ Ω+, 0 < t < T,
u(0, x) = ϕ(x),ut(0, x) = ψ (x) x ∈ Ω+,
u(t, x) = α (t, x) , uxr (t, x) = β (t, x) ,
1 ≤ r ≤ n,0 ≤ t ≤ T, x ∈ S+,∫ x10 ...
∫ xn0 u (t, x) dx1...dxn = ξ(t),0 ≤ t ≤ T,
(1.32)
for the multidimensional hyperbolic partial differential equation. Assume that
ar > a > 0 and f (t, x) ,(t ∈ (0,T) , x ∈ Ω
+), ξ(t), (t ∈ [0,T]), ϕ(x),ψ (x)
(x ∈ Ω
+),
α (t, x) , β (t, x) (t ∈ [0,T] , x ∈ S+) are given smooth functions. Here and in future Ω+ is
the open cube in the n-dimensional Euclidean space Rn (0 < xk < ∞,1 ≤ k ≤ n) with the
boundary S+ and Ω+= Ω+ ∪ S+.
However Laplace transform method described in solving (1.32) can be used only in the case
when (1.32) has constant or polynomial coefficients.
Third, we consider Fourier transform method for solution of the source identification problem
for hyperbolic differential equations.
17
Example 1.2.5 Obtain the Fourier transform solution of the following source identifica-
tion problem
∂2u (t, x)∂t2 −
∂2u (t, x)∂x2 = p (t) e−x2
−(4x2 + 2
)e−t−x2
,
−∞ < x < ∞,0 < t < 1,
u (0, x) = e−x2,ut (0, x) = −e−x2
, x ∈ (−∞,∞) ,∫ ∞−∞
u (t, x) dx = e−t√π, t ≥ 0,
(1.33)
for a one dimensional hyperbolic equation.
Solution. Let us denote
F u (t, x) = u (t, µ) .
Taking the Fourier transform of both sides of the differential equation and initial conditions
(1.33), we can obtainutt (t, µ) + µ2u (t, µ) = p (t) F
e−x2
−e−tF
d2
dx2
(e−x2
),0 < t < 1,
u (0, µ) = Fe−x2
,ut (0, µ) = −F
e−x2
.
(1.34)
Then, we obtain u (t, µ) as the solution of the following Cauchy problemutt (t, µ) + µ2u (t, µ) =
(p (t) + µ2e−t ) F
e−x2,0 < t < 1,
u (0, µ) = Fe−x2
,ut (0, µ) = −F
e−x2
.
(1.35)
Using the D’Alembert’s formula, we obtain
u (t, µ) =eiµt + e−iµt
2F
e−x2
−
eiµt − e−iµt
2iµF
e−x2
+
∫ t
0
eiµ(t−s) − e−iµ(t−s)
2iµ
[(p (s) + µ2e−s
)F
e−x2
]ds
or
u (t, µ) =eiµt + e−iµt
2F
e−x2
−
12
∫ t
−teiµydy
(F
e−x2
)+
12
∫ t
0
∫ t−s
−(t−s)eiµydy
[(p (s) + µ2e−s
)F
e−x2
]ds.
18
Using the formula F f (x ± t) = e±iµtF f (x) ,we obtain
u (t, µ) =12
[F
e−(x+t)2
+ F
e−(x−t)2
]−
12
∫ t
−tF
e−(x+y)
2
dy
+12
∫ t
0
∫ t−s
−(t−s)p (s) F
e−(x+y)
2
dyds +12
∫ t
0
∫ t−s
−(t−s)e−sµ2F
e−(x+y)
2
dyds.
Since µ2Fe−x2
= −F
d2
dx2
(e−x2
),we have that
u (t, µ) =12
[F
e−(x+t)2
+ F
e−(x−t)2
]−
12
∫ t
−tF
e−(x+y)
2
dy
+12
∫ t
0p (s)
∫ t−s
−(t−s)F
e−(x+y)
2
dyds −12
∫ t
0e−s
∫ t−s
−(t−s)F
∂2
∂y2
(e−(x+y)
2)
dyds.
Taking the inverse Fourier transform, we obtain
u (t, x) = F −1 u (t, µ) =12
[e−(x+t)2 + e−(x−t)2
]−
12
∫ t
−te−(x+y)
2dy
+12
∫ t
0p (s)
∫ t−s
−(t−s)e−(x+y)
2dyds −
12
∫ t
0e−s
∫ t−s
−(t−s)
∂2
∂y2
(e−(x+y)
2)
dyds.
Now, applying condition∫ ∞−∞
u (t, x) dx = e−t√π,we obtain
e−t√π =12
[∫ ∞
−∞
e−(x+t)2 dx +∫ ∞
−∞
e−(x−t)2 dx]−
12
∫ t
−t
∫ ∞
−∞
e−(x+y)2dxdy
+12
∫ t
0p (s)
∫ t−s
−(t−s)
∫ ∞
−∞
e−(x+y)2dxdyds −
12
∫ t
0e−s
∫ t−s
−(t−s)
∫ ∞
−∞
∂2
∂y2
(e−(x+y)
2)
dxdyds.
Since∫ ∞−∞
e−x2dx =
√π,we can write
e−t√π =√π −√πt +√π
∫ t
0(t − s) p (s) ds −
12
∫ t
0e−s
∫ t−s
−(t−s)
∫ ∞
−∞
∂2
∂y2
(e−(x+y)
2)
dxdyds.
Since∂2
∂y2
(e−(x+y)
2)= 4 (x + y)2 e−(x+y)
2− 2e−(x+y)
2,
we have that
∫ ∞
−∞
∂2
∂y2
(e−(x+y)
2)
dx = 4∫ ∞
−∞
(x + y)2 e−(x+y)2dx − 2
∫ ∞
−∞
e−(x+y)2dx
19
= 4∫ ∞
−∞
p2e−p2dp − 2
∫ ∞
−∞
e−p2dp = −2
∫ ∞
−∞
pd(e−p2
)− 2√π
= 2∫ ∞
−∞
e−p2dp − 2
√π = 2
√π − 2
√π = 0.
Therefore,
e−t = 1 − t +∫ t
0(t − s) p (s) ds.
From that it follows
−e−t = −1 +∫ t
0p (s) ds.
Taking the derivative, we obtain
e−t = p (t) .
Putting p (t) into the given differential equation (1.35), we obtain the following Cauchy
problem utt (t, µ) + µ2u (t, µ) = e−t (
1 + µ2) F e−x2
,0 < t < 1,
u (0, µ) = Fe−x2
,ut (0, µ) = −F
e−x2
.
We will seek the general solution u (t, µ) of this equation by the following formula
u (t, µ) = uc (t, µ) + up (t, µ) ,
where uc (t, µ) is the solution of homogeneous equation
utt (t, µ) + µ2u (t, µ) = 0,0 < t < 1
and up (t, µ) is the particular solution of nonhomogeneous equation
utt (t, µ) + µ2u (t, µ) = e−t(1 + µ2
)F
e−x2
,0 < t < 1.
Then we have that
uc (t, µ) = c1eiµt + c2e−iµt .
Now, we will seek up (t, µ) by putting the formula up (t, µ) = A (µ) e−t . We have that
A (µ) e−t + µ2 A (µ) e−t = e−t(1 + µ2
)F
e−x2
.
20
From that it follows
A (µ) = Fe−x2
.
Therefore, the general solution of this equation is
u (t, µ) = c1eiµt + c2e−iµt + e−tF
e−x2
.
Using initial conditions, we obtain the system of the equationsu (0, µ) = c1 + c2 + F
e−x2
= F
e−x2
,
ut (0, µ) = iµ (c1 − c2) − Fe−x2
= −F
e−x2
or
c1 + c2 = 0,
c1 − c2 = 0.
Solving this system, we get
c1 = c2 = 0.
Therefore,
u (t, µ) = e−tF
e−x2
.
Taking the inverse Fourier transform, we obtain
u (t, x) = e−te−x2.
So, the exact solution of problem (1.33) is
(u (t, x) , p (t)) =(e−(t+x2), e−t
).
Note that using the same manner one obtain the solution of the following boundary value
problem
∂2u(t, x)∂t2 −
∑|r |=2m
αr∂ |r |u(t, x)∂xr1
1 ...∂xrnn= p(t)q(x) + f (t, x),
0 < t < T, x,r ∈ Rn, |r | = r1 + ... + rn,
u(0, x) = ϕ(x),ut(0, x) = ψ (x) , x ∈ Rn,∫· · ·
∫Rn
u (t, x) dx1...dxn = ξ(t),0 ≤ t ≤ T,
(1.36)
21
for a second order in t and 2m − th order in space variables multidimensional hyper-
bolic differential equation. Assume that αr ≥ α ≥ 0 and f (t, x) , ξ(t), (t ∈ [0,T] , x ∈ Rn) ,
ϕ(x),ψ (x) , (x ∈ Rn) are given smooth functions.
However Fourier transform method described in solving (1.36) can be used only in the case
when (1.36) has constant coefficients. So, all analytical methods described above, namely the
Fourier series method, Laplace transform method and the Fourier transform method can be
used only in the case when the differential equation has constant coefficients. It is well-known
that the most general method for solving partial differential equation with dependent in t and
in the space variables is operator method.
1.3 The Aim of the Thesis
Now, let us briefly describe the contents of the various chapters of the thesis. It consists of
four chapters.
First chapter is the introduction.
Second chapter the theorem on stability of problem (1.1) with local conditions is established.
The first and second order of accuracy difference schemes for the numerical solution of
identification hyperbolic problem (1.1) are presented. The theorems on the stability
estimates for the solution of these difference schemes are established. Numerical results
are provided.
Third chapter the theorem on stability of problem (1.2) with nonlocal conditions is estab-
lished. The first and second order of accuracy difference schemes for the numerical
solution of identification hyperbolic problem (1.2) are presented. The theorems on the
stability estimates for the solution of these difference schemes are proved. Numerical
results are provided.
Fourth chapter contains conclusion.
22
CHAPTER 2
STABILITY OF THE HYPERBOLIC DIFFERENTIAL AND DIFFERENCE
EQUATIONWITH LOCAL CONDITIONS
2.1 Introduction
In this chapter, we consider the source identification problem for a one-dimensional hyperbolic
equation with local conditions
∂2u (t, x)∂t2 −
∂
∂x
(a (x)
∂u (t, x)∂x
)= p (t) q (x) + f (t, x) ,
x ∈ (0, l) , t ∈ (0,T) ,
u (0, x) = ϕ (x) ,ut (0, x) = ψ (x) , x ∈ [0, l] ,
u (t,0) = u (t, l) = 0,∫ l0 u (t, x) dx = ζ (t) , t ∈ [0,T] ,
(2.1)
where u (t, x) and p (t) are unknown functions, a (x) ≥ a > 0, f (t, x) , ζ (t) , ϕ (x) and ψ (x)
are sufficiently smooth functions, and q (x) is a sufficiently smooth function assuming q (0)
= q (l) = 0 and∫ l0 q (x) dx , 0.
2.2 Stability of the Differential Problem (2.1)
To formulate our results, we introduce the Banach space C (H) = C ([0,T] ,H) of all abstract
continuous functions φ (t) defined on [0,T] with values in H, equipped with the norm
‖φ‖C(H) = max06t6T
‖φ (t)‖H .
Let L2 [0, l] be the space of all square-integrable functions γ (x) defined on [0, l] , equipped
with the norm
‖γ‖L2[0,l] =
(∫ l
0|γ (x)|2 dx
) 12,
and let W12 [0, l] ,W
22 [0, l] be Sobolev spaces with norms
‖γ‖W12 [0,l]=
(∫ l
0
[γ2 (x) + γ2
x (x)]
dx) 12,
‖γ‖W22 [0,l]=
(∫ l
0
[γ2 (x) + γ2
xx (x)]
dx) 12,
23
respectively. We introduce the differential operator A defined by the formula
Au(x) = −ddx
(a (x)
du (x)dx
)(2.2)
with the domain
D (A) = u : u,u′′ ∈ L2 [0, l] ,u (0) = u (l) = 0 .
It is easy that A is the self-adjoint positive-definite operator in H = L2 [0, l] . Actually, for all
u, v ∈ L2 [0, l] we have that
〈Au, v〉 =∫ l
0A(u)v (x) dx = −
∫ l
0
ddx
(a (x)
du (x)dx
)v (x) dx
−
(a (x)
du (x)dx
)v(x)
l0+
∫ l
0a (x)
du (x)dx
dv (x)dx
dx
−a (l)du (l)
dxv(l) + a (0)
du (0)dx
v(0) +∫ l
0a (x)
du (x)dx
dv (x)dx
dx
=
∫ l
0a (x)
dv (x)dx
du (x)dx
dx.
〈u, Av〉 =∫ l
0u (x) Av(x)dx = −
∫ l
0u (x)
ddx
(a (x)
dv (x)dx
)dx
− u(x)(a (x)
dv (x)dx
)l0+
∫ l
0a (x)
dv (x)dx
du (x)dx
dx
−u(l)a (l)dv (l)
dx+ u(0)a (0)
dv (0)dx
+
∫ l
0a (x)
dv (x)dx
du (x)dx
dx
=
∫ l
0a (x)
dv (x)dx
du (x)dx
dx.
From that it follows
〈Au, v〉 = 〈u, Av〉
and
〈Au,u〉 =∫ l
0a (x)
du (x)dx
du (x)dx
dx ≥ a∫ l
0
du (x)dx
du (x)dx
dx = a 〈u′,u′〉 . (2.3)
Moreover, using the condition u(0) = 0,we get
u(y) =∫ y
0
du (x)dx
dx =∫ y
0
du (y − t)dt
dt.
24
We will introduce the following function u∗ defined by formula
du∗ (y − t)dt
=
du (y − t)
dt,0 ≤ t ≤ y, y ∈ [0, l] ,
0,otherwise.
Then
u(y) =∫ l
0
du∗ (y − t)dt
dt .
Applying the Minkowsky inequality and the definition of the function u∗ (x), we get(∫ l
0u2 (y) dy
) 12
≤
∫ l
0
(∫ l
0
(du∗ (y − t)
dt
)2dy
) 12
dt
≤
∫ l
0
(∫ l
0
(du (x)
dx
)2dx
) 12
dt
= l
(∫ l
0
(du (x)
dx
)2dx
) 12
.
Therefore,
〈u,u〉 =∫ l
0u2 (y) dy ≤ l2
∫ l
0
(du (x)
dx
)2dxtext (2.4)
= l2∫ l
0
du (x)dx
du (x)dx
dx = l2 〈u′,u′〉 .
Applying inequalities (2.3) and (2.4), we get
〈Au,u〉 ≥al2 〈u,u〉 .
For the self adjoint positive definite operator A we will introduce c(t) and s(t) operator
functions defined by formulas u(t) = c(t)ϕ and v(t) = s(t)ψ,where abstract functions u(t) and
v(t) are solutions of the following Cauchy problems in a Hilbert space H
u′′(t) + Au(t) = 0, t > 0,u(0) = ϕ,u′(0) = 0, (2.5)
v′′(t) + Av(t) = 0, t > 0, v(0) = 0, v′(0) = ψ, (2.6)
respectively. We have the following formulas
c(t) =eit A
12 + e−it A
12
2, s(t) = A−
12
eit A12− e−it A
12
2i
25
and the following estimates hold
‖c(t)‖H→H ≤ 1, A
12 s(t)
H→H
≤ 1. (2.7)
It is based on the spectral represents of unit self adjoint positive definite operator A and
‖ f (A)‖H→H ≤ supδ≤λ<∞
| f (λ)| .
Here f is the bounded function on [δ,∞) .
Moreover, for the differential operator A defined by formula Au = −u′′(x) with the domain
D(A) = u : u(x),u′(x),u′′(x) ∈ E ,we can obtain the following estimates
‖c(t)‖E→E ≤ 1, A
12 s(t)
E→E
≤ 1.
Here, E = Lp(R1) ,1 ≤ p < ∞, Cα
(R1) ,0 ≤ α < 1,R1 = (−∞,∞). The proof of these
estimates is based on the triangle inequality and the following lemma.
Lemma 2.2.1 The following formulas hold:
c(t)ϕ(x) =ϕ(x + t) + ϕ(x − t)
2, (2.8)
s(t)ϕ(x) =12
x+t∫x−t
ϕ(z)dz, (2.9)
A12 s(t)ϕ(x) =
ϕ(x + t) − ϕ(x − t)2i
. (2.10)
Proof. First, we will proof the formula (2.8). Using the definition of operator function c(t),
we can write
u(t, x) = c(t)ϕ(x),
where u(t, x) is the solution of the following Cauchy problem
utt(t, x) − uxx(t, x) = 0, t > 0, x ∈ R1,u(0, x) = ϕ(x),ut(0, x) = 0 (2.11)
for the hyperbolic equation with smooth ϕ(x). Assume that ϕ(±∞) = 0.
Taking the Fourier transform, we get the following Cauchy problem
utt(t, s) + s2u(t, s) = 0, t > 0,u(0, s) = ϕ(x) ,ut(0, s) = 0
26
for the second order differential equation. Taking the Laplace transform, we get
µ2u(µ, s) − µ ϕ(x) + s2u(µ, s) = 0
or
u(µ, s) =µ
µ2 + s2 ϕ(x) .
Sinceµ
µ2 + s2 =12
[1
µ − is+
1µ + is
],
we have that
u(µ, s) =12
[1
µ − is+
1µ + is
]ϕ(x) .
Applying the inverse Laplace transform, we get
u(t, s) =12
[eit ϕ(x) + e−it ϕ(x)
].
Using the shift rule, we get
u(t, s) =12[ϕ(x + t) + ϕ(x − t)] .
Applying the inverse Fourier transform, we get formula (2.8).
Second, we will proof the formula (2.9). Using the definition of operator function s(t),we can
write
u(t, x) = s(t)ψ(x),
where u(t, x) is the solution of the following Cauchy problem
utt(t, x) − uxx(t, x) = 0, t > 0, x ∈ R1,u(0, x) = 0,ut(0, x) = ψ(x) (2.12)
for the hyperbolic equation with smooth ψ(x). Assume that ψ(±∞) = 0.
Taking the Fourier transform, we get the following Cauchy problem
utt(t, s) + s2u(t, s) = 0, t > 0,u(0, s) = 0,ut(0, s) = ψ(x)
for the second order differential equation. Taking the Laplace transform, we get
µ2u(µ, s) − ψ(x) + s2u(µ, s) = 0
27
or
u(µ, s) =1
µ2 + s2 ψ(x) .
Since1
µ2 + s2 =1
2is
[1
µ − is−
1µ + is
],
we have that
u(µ, s) =1
2is
[1
µ − is−
1µ + is
]ψ(x) .
Applying the inverse Laplace transform, we get
u(t, s) =1
2is[eist ψ(x) − e−ist ψ(x)
]=
12
t∫−t
eisydy ψ(x) .
Using the shift rule, we get
u(t, s) =12
t∫
−t
ψ(x + y) dy .
Applying the inverse Fourier transform, we get formula
u(t, x) =12
t∫−t
ψ(x + y)dy.
From that it follows formula (2.9). Lemma 2.2.1 is proved.
Throughout the present thesis, M denotes positive constants, which may differ in time, and
thus are not a subject of precision. However, we will use the notation M(α, β, γ, ...) to stress
the fact that the constant depends only on α, β, γ, ....
We have the following theorem on the stability of problem (2.1):
Theorem 2.2.2 Assume that ϕ ∈ W22 [0, l] ,ψ ∈ W1
2 [0, l] and f (t, x) is a continuously
differentiable function in t and square-integrable in x, and ζ (t) is a twice continuously
differentiable function. Suppose that q (x) is a sufficiently smooth function assuming q (0)
= q (l) = 0 and∫ l0 q (x) dx , 0. Then, for the solution of problem (2.1) the following stability
estimates hold: ∂2u∂t2
C(L2[0,l])
+ ‖u‖C(W22 [0,l])
6 M1 (q)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l](2.13)
28
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
+ ‖ζ ′′‖C[0,T]
],
‖p‖C[0,T] 6 M2 (q)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l]+ ‖ζ ′′‖C[0,T] (2.14)
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
].
Proof. We will use the substitution
u (t, x) = w (t, x) + η (t) q (x) , (2.15)
where η (t) is the function defined by formula
η (t) =∫ t
0(t − s) p (s) ds, η (0) = η′ (0) = 0. (2.16)
It is easy to see that w (t, x) is the solution of the mixed problem
∂2w (t, x)∂t2 −
∂
∂x
(a (x)
∂w (t, x)∂x
)= f (t, x)
+η (t)[
ddx(a (x) q′ (x))
], x ∈ (0, l) , t ∈ (0,T) ,
w (0, x) = ϕ (x) ,wt (0, x) = ψ (x) , x ∈ [0, l] ,
w (t,0) = w (t, l) = 0, t ∈ [0,T] .
(2.17)
Now, we will take an estimate for |p (t)| . Applying the integral overdetermined condition∫ l0 u (t, x) dx = ζ (t) and substitution (2.15), we get
η (t) =ζ (t) −
∫ l0 w (t, x) dx∫ l
0 q (x) dx.
From that and p (t) = η′′ (t), it follows
p (t) =ζ ′′ (t) −
∫ l0∂2
∂t2w (t, x) dx∫ l0 q (x) dx
.
Then, using the Cauchy–Schwarz inequality and the triangle inequality, we obtain
|p (t)| 6|ζ ′′ (t)| +
∫ l0
∂2
∂t2w (t, x) dx∫ l
0 q (x) dx (2.18)
61∫ l
0 q (x) dx|ζ ′′ (t)| +
√l
(∫ l
0
(∂2w (t, x)∂t2
)2
dx
) 12
29
≤ M (q)
[|ζ ′′ (t)| +
∂2w (t, .)∂t2
L2[0,l]
]for all t ∈ [0,T] . From that, it follows
‖p‖C[0,T] 6 M (q)
[‖ζ ′′‖C[0,T] +
∂2w
∂t2
C(L2[0,l])
]. (2.19)
Now, using substitution (2.15), we get
∂2u (t, x)∂t2 =
∂2w (t, x)∂t2 + p (t) q (x) .
Applying the triangle inequality, we obtain ∂2u∂t2
C(L2[0,l])
6
∂2w
∂t2
C(L2[0,l])
+ ‖p‖C[0,T] ‖q‖L2[0,l] . (2.20)
Therefore, the proof of estimates (2.13) and (2.14) is based on equation (2.1), the triangle
inequality, estimates (2.19), (2.20) and on the stability estimate ∂2w
∂t2
C(L2[0,l])
6 M3(q,a)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l](2.21)
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
+ ‖ζ ′′‖C[0,T]
],
for the solution of problem (2.17). This completes the proof of Theorem 2.2.2.
Now, we will prove the estimate (2.21) for the solution of problem (2.17).
Firstly, it is easy to see that problem (2.17) can be written as the abstract Cauchy problem
d2w (t)dt2 + Aw (t) = f (t) − η (t) Aq, t ∈ (0,T) , (2.22)
w (0) = ϕ,w′ (0) = ψ,
in a Hilbert space H with positive-definite self-adjoint operator A defined by formula (2.2).
Here,
w (t) = w(t, x), f (t) = f (t, x)
are unknown and known abstract functions defined on (0,T) with values in H = L2 [0, l] ,
respectively, and ϕ = ϕ(x),ψ = ψ(x),q = q(x) are given.
Secondly, applying the approaches of ( Ashyralyev and Sobolevskii, 2004) , we will prove a
30
lemma that will be needed in the sequel.
Lemma 2.2.3 Assume that ϕ ∈ D(A),ψ,q ∈ D(A12 ), f (t) is a continuously differentiable
abstract function in t with values in H, and η (t) is a twice continuously differentiable function
defined by formula (2.16). Then, for the solution of problem ( 2.22), the following stability
estimate holds d2w (t)dt2
H6 ‖Aϕ‖H +
A12ψ
H
(2.23)
+ ‖ f (0)‖H + T df
dt
C(H)+ M4(q)
∫ t
0|η′′(y)| dy
for any t ∈ [0,T] .
Proof. We have that (see Ashyralyev and Sobolevskii, 2004)
w (t) = c (t) ϕ + s (t)ψ +∫ t
0s (t − y) [ f (y) − η (y) Aq] dy, (2.24)
where
c (t) =eit A
12 + e−it A
12
2and s (t) = A−
12
eit A12− e−it A
12
2i.
Applying equation (2.22) and formula (2.24), we get
d2w (t)dt2 = f (t) − η (t) Aq − Ac (t) ϕ − As (t)ψ
−
∫ t
0As (t − y) [ f (y) − η (y) Aq] dy.
Integrating by parts, we get
d2w (t)dt2 = f (t) − η (t) Aq − Ac (t) ϕ − As (t)ψ − f (t)
+c (t) f (0) + η (t) Aq − c (t) η (0) Aq
+
∫ t
0c (t − y) f ′ (y) dy −
∫ t
0c (t − y) η′ (y) Aqdy.
Therefore,
d2w (t)dt2 = −Ac (t) ϕ − As (t)ψ + c (t) f (0)
+
∫ t
0c (t − y) f ′ (y) dy −
∫ t
0c (t − y) η′ (y) Aqdy.
31
From that it follows
d2w (t)dt2 = −Ac (t) ϕ − As (t)ψ + c (t) f (0) +
∫ t
0c (t − y) f ′ (y) dy
+ [s (t − y) η′ (y) Aq]t0 −∫ t
0s (t − y) η′′ (y) Aqdy
or
d2w (t)dt2 = −Ac (t) ϕ − As (t)ψ + c (t) f (0)
+
∫ t
0c (t − y) f ′ (y) dy −
∫ t
0s (t − y) η′′ (y) Aqdy
=
4∑i=1
Gi (t) ,
where
G1 (t) = c (t) ( f (0) − Aϕ) , G2 (t) = −As (t)ψ,
G3 (t) =∫ t
0c (t − y) f ′ (y) dy, G4 (t) = −
∫ t
0s (t − y) η′′(y)Aqdy
Now, applying the triangle inequality, we obtain d2w (t)dt2
H6
4∑i=1‖Gi (t)‖H
for any t ∈ [0,T] . Therefore, we will estimate ‖Gi (t)‖H , i = 1,2,3,4, separately. Firstly, using
the estimates (2.7), we obtain
‖G1 (t)‖H 6 ‖c (t)‖H→H ‖Aϕ − f (0)‖H 6 ‖Aϕ‖H + ‖ f (0)‖H ,
‖G2 (t)‖H = A
12 A
12 s (t)ψ
H6
A12 s (t)
H→H
A12ψ
H6
A12ψ
H
for any t ∈ [0,T] . Secondly, using the triangle inequality and estimates (2.7), we get
‖G3 (t)‖H ≤∫ t
0‖c (t − y)‖H→H ‖ f ′ (y)‖H dy 6
∫ t
0‖ f ′ (y)‖H dy
≤ t max0≤y≤t
‖ f ′ (y)‖H ≤ T max0≤y≤T
‖ f ′ (y)‖H = T df
dt
C(H)
for any t ∈ [0,T] . Thirdly, using the triangle inequality and estimates (2.7), we get
32
‖G4 (t)‖H ≤∫ t
0
A12 s (t − y)
H→H
A12 q
H|η′′(y)| dy
6
∫ t
0|η′′(y)| dy
A12 q
H≤ M4(q)
∫ t
0|η′′(y)| dy
for any t ∈ [0,T] . Combining these estimates, we obtain estimate (2.23) for the solution of
problem (2.22) for any t ∈ [0,T] .
Theorem 2.2.4 Assume that all conditions of Theorem 2.2.1 are satisfied. Then, for the
solution of problem (2.17) the stability estimate (2.21) holds.
Proof. Putting H = L2 [0, l] , ϕ = ϕ(x),ψ = ψ(x),q = q(x), f (t) = f (t, x) ,w (t) = w (t, x)
and applying estimates (2.18), (2.23), we get ∂2w (t)∂t2
L2[0,l]
6 M5 (a)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l]
]+ ‖ f (0, ·)‖L2[0,l] + T
∂ f∂t
C(L2[0,l])
+ M (q)M4 (q)
×
[∫ t
0|ζ ′′(y)| dy +
∫ t
0
∂2w (y, ·)
∂y2
L2[0,l]
dy
]for any t ∈ [0,T] . By the Gronwall’s inequality, we conclude that, for any t ∈ [0,T] , the
following estimate for the solution of problem (2.17) holds: ∂2w (t)∂t2
L2[0,l]
6M5 (a)
[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l]
]+ ‖ f (0, ·)‖L2[0,l]
+ T
[ ∂ f∂t
C(L2[0,l])
+ M (q)M4 (q) ‖ζ ′′‖C[0,T]
]eM(q)M4(q)t .
This completes the proof of Theorem 2.2.4.
2.3 Stability of Difference Scheme
To formulate our results on a difference problem, we introduce the Banach space Cτ (H) =
C ([0,T]τ ,H) of all abstract grid functions φτ = φ (tk)Nk=0 defined on
[0,T]τ = tk = kτ,0 6 k 6 N,Nτ = T ,
with values in H, equipped with the norm
33
‖φτ‖Cτ(H) = max06k6N
‖φ (tk)‖H .
Moreover, L2h = L2 [0, l]h is the Hilbert space of all grid functions γh (x) = γnMn=0 defined
on
[0, l]h = xn = nh,0 6 n 6 M,Mh = l ,
equipped with the norm γh
L2h=
M∑
i=0|γi |
2 h
12
,
and W12h = W1
2 [0, l]h ,W22h = W2
2 [0, l]h are the discrete analogues of Sobolev spaces of all
grid functions γh (x) = γnMn=0 defined on [0, l]h with norms
γh
W12h=
M∑
i=0|γi |
2 h +M∑
i=1
γi − γi−1h
2 h
12
,
γh
W22h=
M∑
i=0|γi |
2 h +M−1∑i=1
γi+1 − 2γi + γi−1
h2
2 h
12
,
respectively. For the differential operator A defined by (2.2), we introduce the difference
operator Ah defined by formula
Ahϕh (x) =
−
1h
(a (xn+1)
ϕn+1 − ϕn
h− a (xn)
ϕn − ϕn−1h
)M−1
n=1,
acting in the space of grid functions ϕh (x) = ϕnMn=0 defined on [0, l]h, satisfying the
conditions ϕM = ϕ0 = 0.
It is easy that Ah is the self-adjoint positive-definite operator in H = L2h = L2 [0, l]h . Actually,
we have that ⟨Ahuh, vh⟩ = ∑
x∈[0,l]h
Ahuh(x)vh(x)h
= −
M−1∑n=1
1h
(a (xn+1)
un+1 − un
h− a (xn)
un − un−1h
)vnh
= −
M−1∑n=1
a (xn+1)un+1 − un
hvn +
M−1∑n=1
a (xn)un − un−1
hvn
= −
M∑n=2
a (xn)un − un−1
hvn−1 +
M−1∑n=1
a (xn)un − un−1
hvn
34
= −a (xM)uM − uM−1
hvM−1 −
M−1∑n=1
a (xn)un − un−1
hvn−1
+
M−1∑n=1
a (xn)un − un−1
hvn
=1h
a (xM) uM−1vM−1 +
M−1∑n=1
a (xn)un − un−1
hvn − vn−1
hh
= a (xM)uM − uM−1
hvM − vM−1
hh +
M−1∑n=1
a (xn)un − un−1
hvn − vn−1
hh
=
M∑n=1
a (xn)un − un−1
hvn − vn−1
hh,
⟨uh, Ahv
h⟩ = ∑x∈[0,l]h
uh(x)Ahvh(x)h
= −
M−1∑n=1
un1h
(a (xn+1)
vn+1 − vn
h− a (xn)
vn − vn−1h
)h
= −
M−1∑n=1
una (xn+1)vn+1 − vn
h+
M−1∑n=1
una (xn)vn − vn−1
h
= −
M∑n=2
un−1a (xn)vn − vn−1
h+
M−1∑n=1
una (xn)vn − vn−1
h
= −uM−1a (xM)vM − vM−1
h−
M−1∑n=1
a (xn)vn − vn−1
hun−1
+
M−1∑n=1
a (xn) unvn − vn−1
h
=1h
a (xM) uM−1vM−1 +
M−1∑n=1
a (xn)un − un−1
hvn − vn−1
hh
= a (xM)uM − uM−1
hvM − vM−1
hh +
M−1∑n=1
a (xn)un − un−1
hvn − vn−1
hh
=
M∑n=1
a (xn)un − un−1
hvn − vn−1
hh,
From that it follows ⟨Ahuh, vh⟩ = ⟨
uh, Ahvh⟩35
and ⟨Ahuh,uh⟩ = M∑
n=1a (xn)
un − un−1h
un − un−1h
h (2.25)
≥ aM∑
n=1
un − un−1h
un − un−1h
h = a⟨Dhuh,Dhuh⟩ .
Here
Dhuh(x) =un − un−1
h
M
n=1.
Moreover, using the condition u0 = 0,we get
um =
m∑n=1
un − un−1h
h =m∑
i=1
um−i+1 − um−i
hh.
We will introduce the mesh function uh∗ (x) defined by the following formula
(um−i+1 − um−i
h
)∗=
um−i+1 − um−i
h,1 ≤ i ≤ m,1 ≤ m ≤ M,
0,otherwise.
Then
um =
M∑i=1
(um−i+1 − um−i
h
)∗
h.
Applying the discrete analogue of Minkowsky inequality and the definition of the mesh
function uh∗ (x), we get(M−1∑m=1
u2mh
) 12
≤
M∑i=1
(M−1∑m=1
(um−i+1 − um−i
h
)2
∗h
) 12
h
≤
M∑i=1
h
(M∑
n=1
(un − un−1h
)2h
) 12
= Mh
(M∑
n=1
(un − un−1h
)2h
) 12
= l
(M∑
n=1
(un − un−1h
)2h
) 12
.
Therefore, ⟨uh,uh⟩ = M−1∑
m=1u2
mh ≤ l2M∑
n=1
(un − un−1h
)2h (2.26)
= l2M∑
n=1
un − un−1h
un − un−1h
h = l2 ⟨Dhuh,Dhuh⟩ .
36
Applying inequalities (2.25) and (2.26), we get⟨Ahuh,uh⟩ ≥ a
l2
⟨uh,uh⟩ .
2.3.1 The first order of accuracy difference scheme
For the numerical solution
uknN
k=0
M
n=0of problem (2.1), we consider the first order of
accuracy difference scheme
uk+1n − 2uk
n + uk−1n
τ2 −
(a (xn+1)
uk+1n+1 − uk+1
n
h2 − a (xn)uk+1
n − uk+1n−1
h2
)= pk q (xn) + f (tk, xn) ,
tk = kτ, xn = nh,1 6 k 6 N − 1,1 6 n 6 M − 1,Nτ = T,
u0n = ϕ (xn) ,
u1n − u0
n
τ= ψ (xn) ,0 6 n 6 M,Mh = l,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = ζ (tk+1) ,−1 6 k 6 N − 1.
(2.27)
Here, it is assumed that qM = q0 = 0, and∑M−1
i=1 qi , 0.We have the following theorem on
the stability of the difference scheme (2.27):
Theorem 2.3.1 For the solution of difference scheme (2.27), the following stability esti-
mates hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1
N−1k=1
Cτ(W2
2h)(2.28)
6 M6 (q)[ ϕh
W2
2h+
ψh
W12h+
f h1
L2h
+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M7 (q)[ ϕh
W2
2h+
ψh
W22h+
f h1
L2h(2.29)
+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.Here and throughout this subsection f h
k (x) = f (tk, xn)Mn=0 ,1 ≤ k ≤ N − 1.
37
Proof. We will use the substitution
ukn = wk
n + ηk qn, (2.30)
where
qn = q (xn) ,
and
ηk+1 =
k∑i=1(k + 1 − i) piτ
2,1 6 k 6 N − 1, η0 = η1 = 0. (2.31)
It is easy to see thatwk
nN
k=0
M
n=0is the solution of the difference problem
wk+1n − 2wk
n + wk−1n
τ2 −1h
(a (xn+1)
wk+1n+1 − w
k+1n
h− a (xn)
wk+1n − wk+1
n−1h
)= f (tk, xn) +
1h
[a (xn+1)
qn+1 − qn
h− a (xn)
qn − qn−1h
]ηk+1,
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = ϕ (xn) ,
w1n − w
0n
τ= ψ (xn) ,0 6 n 6 M,
wk+10 = wk+1
M = 0,−1 6 k 6 N − 1.
(2.32)
Now, we will take an estimate for |pk | .Using the overdetermined condition∑M−1
i=1 uk+1i h =
ζ (tk+1) and substitution (2.30), one can obtain
ηk+1 =ζk+1 −
∑M−1i=1 wk+1
i h∑M−1i=1 qih
. (2.33)
Then, using the formulas pk =ηk+1−2ηk+ηk−1
τ2 and (2.33), we get
pk =ζk+1 − 2ζk + ζk−1 −
∑M−1i=1
(wk+1
i − 2wki + w
k−1i
)h
τ2 ∑M−1i=1 qih
.
Then, applying the discrete analogue of the Cauchy–Schwartz inequality and the triangle
inequality, we obtain
|pk | 61∑M−1
i=1 qih (2.34)
×
[ ζk+1 − 2ζk + ζk−1
τ2
+ M−1∑i=1
wk+1i − 2wk
i + wk−1i
τ2
h
]6
1∑M−1i=1 qih
ζk+1 − 2ζk + ζk−1
τ2
+ √l
wk+1
i − 2wki + w
k−1i
τ2
M
i=0
L2h
38
6 M8 (q)
[ ζk+1 − 2ζk + ζk−1
τ2
+ whk+1 − 2wh
k + whk−1
τ2
L2h
]for all 1 6 k 6 N − 1. From that, it follows
pkN−1k=1
C[0,T]τ
6 M8 (q) ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
(2.35)
+
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
.Now, using substitution (2.30), we get
uk+1n − 2uk
n + uk−1n
τ2 =wk+1
n − 2wkn + w
k−1n
τ2 + pk q (xn) .
Applying the triangle inequality, we obtain
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
6
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
(2.36)
+ pk
N−1k=1
C[0,T]τ
q (xn)Mn=0
L2h
.
Therefore, the proof of estimates (2.28) and (2.29) is based on equation (2.27), the triangle
inequality, estimates (2.35), (2.36) and on the following stability estimate for the solution of
difference problem (2.32): wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
6 M10(q)[ ϕh
W2
2h+
ψh
W12h
(2.37)
+ f h
1
L2h+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.This completes the proof of Theorem 3.2.1.
Now, we will prove estimate (2.37) for the solution of difference problem (2.32). Firstly, it
is easy to see that difference scheme (2.32) can be written as the abstract Cauchy difference
problem wk+1 − 2wk + wk−1
τ2 + Awk+1 = fk − ηk+1 Aq,1 6 k 6 N − 1,
w0 = ϕ,w1 − w0
τ= ψ
(2.38)
39
in a Hilbert space H = L2h with positive-definite self-adjoint operator A. Here,
wkNk=0 =
wh
k
Nk=0 , fkN−1
k=1 =
f hk
N−1k=1
are unknown and known abstract mesh functions defined on [0,T]τ with values in H = L2h,
respectively, and ϕ = ϕh,ψ = ψh,q = qh are given elements.
Secondly, applying the approaches of (Ashyralyev and Sobolevskii, 2004) we will prove a
lemma that will be needed in the sequel.
Lemma 2.3.2 Assume that ϕ ∈ D(A),ψ,q ∈ D(A12 ). Then, for the solution of difference
problem (2.38), the following stability estimate holds for any 1 6 k 6 N − 1: wk+1 − 2wk + wk−1
τ2
H6 ‖Aϕ‖H +
A12ψ
H
+ ‖ f1‖H (2.39)
+ |η2 | ‖Aq‖H + T
fk − fk−1τ
N
k=2
C(H)
+
k∑s=1
ηs+1 − 2ηs + ηs−1
τ2
τ A12 q
H
Proof. It is clear that there exists a unique solution of this initial value problem, and for the
solution of (2.38), the following formula is satisfied (see Ashyralyev and Sobolevskii, 2004)
w0 = ϕ, w1 = ϕ + τψ,
wk =12[Rk−1 + Rk−1]ϕ + τ(R − R)−1
[Rk − Rk
]ψ (2.40)
+
k−1∑s=1
RR(R − R
)−1 [Rk−s − Rk−s
] fs − ηs+1 Aq τ2,
2 ≤ k ≤ N,
where R =(I + iτA
12
)−1and R =
(I − iτA
12
)−1.Using the spectral property of the self-adjoint
positive-definite operator, we get
‖R‖H→H ≤ 1, τA
12 R
H→H
≤ 1. (2.41) R
H→H≤ 1,
τA12 R
H→H
≤ 1. (2.42)
40
Now, we will establish estimates for wk+1−2wk+wk−1
τ2
H,1 ≤ k ≤ N − 1. Applying equation
(2.38) and formula (2.40), we get
wk+1 − 2wk + wk−1
τ2 (2.43)
= fk − ηk+1 Aq −12[Rk + Rk]Aϕ − τA(R − R)−1
[Rk+1 − Rk+1
]ψ
−
k∑s=1
ARR(R − R
)−1×
[Rk+1−s − Rk+1−s
] fs − ηs+1 Aq τ2
. = fk − ηk+1 Aq −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ
+
(τA
12
2i
) (iτA
12
)−1
k∑s=1
Rk−s (I − R) +k∑
s=1Rk−s
(I − R
) fs − ηs+1 Aq
= fk − ηk+1 Aq −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ
−12
k∑
s=1
[Rk−s − Rk+1−s] + k∑
s=1
[Rk−s − Rk+1−s
]( fs − ηs+1 Aq)
= fk − ηk+1 Aq −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ (2.44)
−12
k∑
s=1
[Rk−s + Rk−s
]( fs − ηs+1 Aq) −
k∑s=1
[Rk+1−s − Rk+1−s
]( fs − ηs+1 Aq)
2 ≤ k ≤ N .
Applying Abel’s formula to (2.44), we can write
wk+1 − 2wk + wk−1
τ2
= fk − ηk+1 Aq −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ
−12
k+1∑s=2
[Rk+1−s + Rk+1−s
]−
k∑s=1
[Rk+1−s + Rk+1−s
]( fs−1 − ηs Aq)
= fk − ηk+1 Aq −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ
−12
k∑
s=2
[Rk+1−s + Rk+1−s
]( fs−1 − ηs Aq) + 2 ( fk − ηk+1 Aq)
−
k∑s=2
[Rk+1−s + Rk+1−s
]( fs − ηs+1 Aq) −
[Rk + Rk
]( f1 − η2 Aq)
41
= −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − Rk R−1
]A
12ψ
−12
k∑s=2
[Rk+1−s + Rk+1−s
]( fs−1 − fs) +
12
[Rk + Rk
]f1
+12
k∑s=2
[Rk+1−s + Rk+1−s
](ηs − ηs+1) Aq −
12
[Rk + Rk
]η2 Aq,
Now, we have that
wk+1 − 2wk + wk−1
τ2
= −12[Rk + Rk]Aϕ −
12i
[Rk R−1 − R−1Rk
]A
12ψ
+12
[Rk + Rk
]f1 −
12
k∑s=2
[Rk+1−s − Rk+1−s
]( fs−1 − fs)
+12
k∑s=1
[Rk+1−s − Rk+1−s
](ηs − ηs+1) Aq
=
4∑i=1
Gi (k) ,2 ≤ k ≤ N,
where
G1 (k) = −12[Rk + Rk]Aϕ +
12[Rk + Rk] f1,
G2 (k) = −12i
[Rk R−1 − R−1Rk
]A
12ψ,
G3 (k) = −12
k∑s=2
[Rk+1−s + Rk+1−s
]( fs−1 − fs) ,
G4 (k) =12
k∑s=1
[Rk+1−s + Rk+1−s
](ηs − ηs+1) Aq.
Now, applying the triangle inequality, we obtain
‖G1 (k)‖H 612
[ Rk
H→H +
Rk
H→H
][‖Aϕ‖H + ‖ f1‖H]
≤ ‖Aϕ‖H + ‖ f1‖H ,
‖G2 (k)‖H 612
[ Rk R−1
H→H+
R−1Rk
H→H
] A12ψ
H
≤
A12ψ
H
42
for any 1 ≤ k ≤ N − 1. Secondly, using the triangle inequality and estimates (2.41), (2.42),
we get
‖G3 (k)‖H 6k∑
s=2
12
[ Rk+1−s
H→H +
Rk+1−s
H→H
]‖ fs−1 − fs‖H
6k∑
s=2‖ fs−1 − fs‖H ≤ T
fk − fk−1τ
N−1
k=1
Cτ(H)
for any 1 ≤ k ≤ N − 1. Thirdly, applying Abel’s formula to the sum G4 (k), we can write
G4 (k) =12
(iτA
12
)−1
k∑s=1
Rk−s (I − R) −k∑
s=1Rk−s
(I − R
)(ηs − ηs+1) Aq
=12
(iτA
12
)−1
k∑s=1
[Rk−s − Rk−s
](ηs − ηs+1) Aq
−
k∑s=1
[Rk+1−s − Rk+1−s
](ηs − ηs+1) Aq
=12
(iτA
12
)−1
k+1∑s=2
[Rk+1−s − Rk+1−s
](ηs−1 − ηs) Aq
−
k∑s=1
[Rk+1−s − Rk+1−s
](ηs − ηs+1) Aq
=12
(iτA
12
)−1
k∑s=1
[Rk+1−s − Rk+1−s
](ηs−1 − ηs) Aq
−
[Rk − Rk
](η0 − η1) Aq
−
k∑s=1
[Rk+1−s − Rk+1−s
](ηs − ηs+1) Aq
=
12
(iτA
12
)−1
k∑s=1
[Rk+1−s − Rk+1−s
](ηs−1 − 2ηs − ηs+1) Aq
=
(iτA
12
)−1 12
k∑
s=1
[Rk+1−s − Rk+1−s
] (ηs+1 − 2ηs + ηs−1
τ2
)τ2 Aq
Finally, using the triangle inequality and estimates (2.41), (2.42), we get
43
‖G4 (k)‖H 6k∑
s=1
12
[ Rk+1−s
H→H +
Rk+1−s
H→H
] ηs+1 − 2ηs + ηs−1
τ2
τ A12 q
H
6k∑
s=1
ηs+1 − 2ηs + ηs−1
τ2
τ A12 q
H
for any 1 ≤ k ≤ N − 1. Combining these estimates, we obtain estimate ( 2.39) for the solution
of problem (2.38) for any 1 ≤ k ≤ N − 1.
Theorem 2.3.3 For the solution of difference problem (2.32), the stability estimate ( 2.38)
holds.
Proof. Putting H = L2h, ϕ = ϕh,ψ = ψh,q = qh, Aϕ = Ahϕh,wk = wh
k , fk = f hk and
applying estimates (3.35) and (2.39), we get whk+1 − 2wh
k + whk−1
τ2
L2h
6 M12 (q)[ ϕh
W2
2h+
ψh
W12h
]+
f h1
L2h+ T
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+ M13 (q)M8 (q)
×
k∑s=1
[ ζs+1 − 2ζs + ζs−1
τ2
+ whs+1 − 2wh
s + whs−1
τ2
L2h
]τ
for any 1 ≤ k ≤ N − 1. By the difference analogue of Gronwall’s inequality, we conclude that whk+1 − 2wh
k + whk−1
τ2
L2h
61
1 − M13 (q)M8 (q) τ
M12 (q)
[ ϕh
W22h+
ψh
W12h
]+
f h1
L2h+ T
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+M13 (q)M8 (q)T
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
e(k−1)τ M13(q)M8(q)1−M13(q)M8(q)τ
for any 1 ≤ k ≤ N − 1.This completes the proof of Theorem 2.3.3.
44
2.3.2 The second order of accuracy difference scheme
For the numerical solution
uknN
k=0
M
n=0of problem (2.1), we consider the second order of
accuracy difference scheme
uk+1n − 2uk
n + uk−1n
τ2 −12h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)−
14h
(a (xn+1)
uk+1n+1 − uk+1
n + uk−1n+1 − uk−1
n
h
−a (xn)uk+1
n − uk+1n−1 + uk−1
n − uk−1n−1
h
)= pk qn + f (tk, xn) ,
Nτ = T,1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h
)= ψ (xn) +
τ
2f (0, xn)
+τ
2
[1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ p0qn
],
1 6 n 6 M − 1,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = ζ (tk+1) ,−1 6 k 6 N − 1
(2.45)
Here, it is assumed that qM = q0 = 0, and∑M−1
i=1 qi , 0.We have the following theorem on
the stability of the difference scheme (2.45):
Theorem 2.3.4 For the solution of difference scheme (2.45), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1 + 2uh
k + uhk−1
4
N−1
k=1
Cτ(W2
2h)
(2.46)
6 M14 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M15 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h(2.47)
45
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.Here and throughout this subsection f h
k (x) = f (tk, xn)Mn=0 ,1 ≤ k ≤ N − 1.
Proof. We will use the substitution
ukn = wk
n + ηk qn, (2.48)
where
ηk =
k∑i=1
(k − i) pi + (k − (i − 1)) pi−1
2
τ2,1 6 k 6 N, η0 = 0, (2.49)
It is easy to see thatwk
nN
k=0
M
n=0is the solution of the difference problem
wk+1n − 2wk
n + wk−1n
τ2 −12h
(a (xn+1)
wkn+1 − w
kn
h− a (xn)
wkn − w
kn−1
h
)−
14h
(a (xn+1)
wk+1n+1 − w
k+1n
h− a (xn)
wk+1n − wk+1
n−1h
)−
14h
(a (xn+1)
wk−1n+1 − w
k−1n
h− a (xn)
wk−1n − wk−1
n−1h
)= f (tk, xn)
+1h
[a (xn+1)
qn+1 − qn
h− a (xn)
qn − qn−1h
]×
14(ηk+1 + 2ηk + ηk−1) ,
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = ϕ (xn) ,0 6 n 6 M,
w1n − w
0n
τ−τ
h
(a (xn+1)
w1n+1 − w
0n+1 − w
1n + w
0n
h
−a (xn)w1
n − w0n − w
1n−1 + w
0n−1
h
)−τ
h
(a (xn+1)
qn+1 − qn
h
−a (xn)qn − qn−1
h
)η1 = ψ (xn) +
τ
2f (0, xn)
+τ
2
[1h
(a (xn+1)
w0n+1 − w
0n
h− a (xn)
w0n − w
0n−1
h
)],
1 6 n 6 M − 1,
wk+10 = wk+1
M = 0,−1 6 k 6 N − 1.
(2.50)
Now, we will take an estimate for |pk | .Using the overdetermined condition∑M−1
i=1 uk+1i h =
ζ (tk+1) and substitution (2.48), one can obtain
46
ηk =ζk −
∑M−1i=1 wk
i h∑M−1i=1 qih
. (2.51)
Then, using the formulas pk =ηk+1−2ηk+ηk−1
τ2 and (2.51), we get
pk =ζk+1 − 2ζk + ζk−1 −
∑M−1i=1
(wk+1
i − 2wki + w
k−1i
)h
τ2 ∑M−1i=1 qih
.
Then, applying the discrete analogue of the Cauchy–Schwarz inequality and the triangle
inequality, we obtain
|pk | 61∑M−1
i=1 qih ×
[ ζk+1 − 2ζk + ζk−1
τ2
+ M−1∑i=1
wk+1i − 2wk
i + wk−1i
τ2
h
](2.52)
61∑M−1
i=1 qih
ζk+1 − 2ζk + ζk−1
τ2
+ √l
wk+1
i − 2wki + w
k−1i
τ2
M
i=0
L2h
6 M16 (q)
[ ζk+1 − 2ζk + ζk−1
τ2
+ whk+1 − 2wh
k + whk−1
τ2
L2h
]for all 1 6 k 6 N − 1. From that it follows pk
N−1k=1
C[0,T]τ
6 M16 (q) ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
(2.53)
+
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
.Now, using substitution (2.48), we get
uk+1n − 2uk
n + uk−1n
τ2 =wk+1
n − 2wkn + w
k−1n
τ2 + pk q (xn) .
Applying the triangle inequality, we obtain
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
6
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
(2.54)
+ pk
N−1k=1
C[0,T]τ
q (xn)Mn=0
L2h
.
Therefore, the proof of estimates (2.46) and (2.47) is based on equation (2.45), the triangle
inequality, estimates (2.53), (2.54) and on the following stability estimate for the solution of
difference problem (2.50):
47
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
6 M17(q)[ ϕh
W2
2h+
ψh
W12h
(2.55)
+ f h
0
L2h+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.
.This completes the proof of Theorem 2.3.4.
Now, we will prove estimate (2.55) for the solution of difference problem (2.50). Firstly, it
is easy to see that difference scheme (2.50) can be written as the abstract Cauchy difference
problem
wk+1 − 2wk + wk−1
τ2 + 14 A (wk+1 + 2wk + wk−1) = θk,
θk = fk − 14 (ηk+1 + 2ηk + ηk−1) Aq,
θk = θ(tk), tk = kτ,1 6 k 6 N − 1,Nτ = 1,(I + τ2 A
)τ−1 (w1 − w0) =
τ2 (θ0 − Aw0) + ψ,
θ0 = f (0) + p0q,w0 = ϕ,
(2.56)
in a Hilbert space H = L2h with positive-definite self-adjoint operator A. Here,
wkNk=0 =
wh
k
Nk=0 , fkN−1
k=1 =
f hk
N−1k=1
are unknown and known abstract mesh functions defined on [0,T]τ with values in H = L2h,
respectively, and ϕ = ϕh,ψ = ψh,q = qh are given elements.
Secondly, applying the approaches of ( Ashyralyev and Sobolevskii, 2004) ) we will prove a
lemma that will be needed in the sequel.
Lemma 2.3.5 Assume that ϕ ∈ D(A),ψ,q ∈ D(A12 ). Then, for the solution of difference
problem (2.56), the following stability estimate holds for any 1 6 k 6 N − 1: wk+1 − 2wk + wk−1
τ2
H6 M (q)
[‖Aϕ‖H +
A12ψ
H+ ‖ f0‖H + |η1 | (2.57)
+
fk − fk−1τ
N
k=1
C(H)
+
k∑s=1
ηs+1 − 2ηs + ηs−1
τ2
τfor any 1 6 k 6 N − 1.
48
Proof. It is clear that there exists a unique solution of this initial value problem, and for the
solution of (2.56), the following formula is satisfied (see Ashyralyev and Sobolevskii, 2004)
w0 = ϕ,w1 =(I + τ2 A
)−1[(
I +τ2
2A)ϕ + τψ +
τ2
2θ0
](2.58)
wk =
[Rk +
12i
A−12
(I −
iτA12
2
) [Rk − Rk
]×
((I +
iτA12
2
)τ
2A − iA
12
(I + τ2 A
)) (I + τ2 A
)−1]ϕ
+i2
A−12
(I −
iτA12
2
) [Rk − Rk
] (I + τ2 A
)−1(I +
iτA12
2
)ψ
+i2
A−12
(I −
iτA12
2
) [Rk − Rk
] (I + τ2 A
)−1(I +
iτA12
2
)τ
2θ0
−
k−1∑s=1
τ
2iA−
12
[Rk−s − Rk−s
]θs,
2 ≤ k ≤ N .
Applying Abel’s formula, we can write
wk =
Rk −
12
iA−12 B
[Rk − Rk
] (Bτ
2A − iA
12
(I + τ2 A
)) (I + τ2 A
)−1ϕ (2.59)
+
i2
A−12 BB
[Rk − Rk
] (I + τ2 A
)−1ψ
+
i2
A−12 BB
[Rk − Rk
] (I + τ2 A
)−1τ
2θ0
+A−1
2
k−1∑s=2
[BRk−s − BRk−s
](θs−1 − θs)
,
+A−1
2
[B + B
]θk−1 −
[BRk−1 − BRk−1
]θ1
,
2 ≤ k ≤ N,
where
R = BB−1 =
(I −
iτA12
2
) (I +
iτA12
2
)−1
R = BB−1 =
(I +
iτA12
2
) (I −
iτA12
2
)−1
.
49
Using the spectral property of the self-adjoint positive-definite operator, we get
‖R‖H→H ≤ 1, R
H→H
≤ 1,
(I ±
iτA12
2
)−1
H→H
≤ 1, (2.60)
(I ± iτA12
)−1
H→H≤ 1,
τA12
(I ± iτA
12
)−1
H→H≤ 1. (2.61)
Now, we will establish estimates for wk+1−2wk+wk−1
τ2
H,1 ≤ k ≤ N − 1. Applying equation (
2.56), formula (2.59) and identities
I + R = 2B−1, I + R = 2B−1, (2.62)
we get
Awk+1 + wk + wk−1
4(2.63)
=
B−1Rk B−1 −
14
iτA12
[BRk−1B−1 − B−1Rk−1B
]×
(I + τ2 A
)−1−
12
[B−1Rk − B−1Rk−1
]Aϕ
+12
[BRk−1B−1 − B−1Rk−1B
] (I + τ2 A
)−1
iA12ψ
+14
iτA
12
[BRk−1B−1 − B−1Rk−1B
] (I + τ2 A
)−1θ0
+12
k−1∑s=2
[B−1Rk−s + B−1Rk−s
](θs−1 − θs) +
14
[τ2 AB−1B−1
]θk
+12
[B−1 + B−1
]θk−1 −
12
[B−1Rk−1 + B−1Rk−1
]θ1
=
4∑i=1
Ji (k) ,
where
J1 (k) =
B−1Rk B−1 −14
iτA12
[BRk−1B−1 − B−1Rk−1B
] (I + τ2 A
)−1
−12
[B−1Rk − B−1Rk−1
]Aϕ,
J2 (k) =12
[BRk−1B−1 − B−1Rk−1B
] (I + τ2 A
)−1
iA12ψ,
J3 (k) =14
iτA
12
[BRk−1B−1 − B−1Rk−1B
] (I + τ2 A
)−1θ0
50
+14
[τ2 AB−1B−1
]θk +
12
[B−1 + B−1
]θk−1
−12
[B−1Rk−1 + B−1Rk−1
]θ1,
J4 (k) =12
k−1∑s=2
[B−1Rk−s + B−1Rk−s
](θs−1 − θs) .
Now, applying the triangle inequality, we obtain Awk+1 + wk + wk−1
4
H6
4∑i=1‖Ji (k)‖H (2.64)
for any 1 ≤ k ≤ N − 1. Therefore, we will estimate ‖Ji (k)‖H , i = 1,2,3,4, separately. Firstly,
using estimates (2.60), (2.61), we get
‖J1 (k)‖H 6 B−1Rk B−1
H→H
+14
[ iτA12 BRk−1B−1
(I + τ2 A
)−1
H→H
+
iτA12 B−1Rk−1B
(I + τ2 A
)−1
H→H
]+
12
[ B−1Rk
H→H+
B−1Rk−1
H→H
]‖Aϕ‖H
≤ M ‖Aϕ‖H ,
‖J2 (k)‖H 612
[ BRk−1B−1(I + τ2 A
)−1
H→H
+
B−1Rk−1B(I + τ2 A
)−1
H→H
A12ψ
H
]≤ M
A12ψ
H
for any 1 ≤ k ≤ N − 1. Secondly, using the triangle inequality and estimates (2.60), (2.61),
we get
‖J3 (k)‖H 614
[ BRk−1B−1τA12
(I + τ2 A
)−1
H→H
+
B−1Rk−1BτA12
(I + τ2 A
)−1
H→H
]‖θ0‖H
+14
[ τ2 AB−1B−1
H→H‖θk ‖H
]+
12
[ B−1 H→H +
B−1
H→H
]‖θk−1‖H
+12
[ B−1Rk−1
H→H+
B−1Rk−1
H→H
]‖θ1‖H
51
≤114
[‖θ0‖H +
k∑i=1‖θi − θi−1‖H
],
‖J4 (k)‖H 6k−1∑s=2
12
[ B−1Rk−s
H→H+
B−1Rk−s
H→H
]‖θs−1 − θs‖H
6k−1∑s=2‖θs−1 − θs‖H
for any 1 ≤ k ≤ N − 1. Using the triangle inequality and estimate (2.64), we get
A wk+1 + 2wk + wk−1
4
H≤ M ‖Aϕ‖H + M
A12ψ
H+
k−1∑s=2‖θs−1 − θs‖H (2.65)
+114
[‖θ0‖H +
k∑i=1‖θi − θi−1‖H
]for any 1 ≤ k ≤ N − 1.Using the triangle inequality and estimate, we get wk+1 − 2wk + wk−1
τ2
H≤ A
wk+1 + 2wk + wk−14
H+ ‖θk ‖H
≤ M ‖Aϕ‖H + M A
12ψ
H+
k−1∑s=2‖θs−1 − θs‖H
+154
[‖θ0‖H +
k∑i=1‖θi − θi−1‖H
]for any 1 ≤ k ≤ N − 1. Combining these estimates, we obtain estimate ( 2.57) for the solution
of difference problem (2.56) for any 1 ≤ k ≤ N − 1.
Theorem 2.3.6 For the solution of difference problem (2.50), the stability estimate (2.56)
holds.
Proof. Putting H = L2h, ϕ = ϕh,ψ = ψh,q = qh, Aϕ = Ahϕh,wk = wh
k , fk = f hk and
applying estimates (2.52) and (2.57), we get whk+1 − 2wh
k + whk−1
τ2
L2h
6 M20 (q)[ ϕh
W2
2h+
ψh
W12h
]
+ f h
0
L2h+ T
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+ M21 (q)M16 (q)
×
k∑s=1
[ ζs+1 − 2ζs + ζs−1
τ2
+ whs+1 − 2wh
s + whs−1
τ2
L2h
]τ
52
for any 1 ≤ k ≤ N − 1. By the difference analogue of Gronwall’s inequality, we conclude that whk+1 − 2wh
k + whk−1
τ2
L2h
61
1 − M21 (q)M16 (q) τ
M20 (q)
[ ϕh
W22h+
ψh
W12h
]+
f h0
L2h+ T
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+ M21 (q)M16 (q)T
×
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
e(k−1)τ M21(q)M16(q)1−M21(q)M16(q)τ
for any 1 ≤ k ≤ N − 1.This completes the proof of Theorem 2.3.6.
Applying same approach with this work we can study another the second order of accuracy
difference scheme for the numerical solution of problem (2.1).
uk+1n − 2uk
n + uk−1n
τ2 −12h
(a (xn+1)
uk+1n+1 − uk+1
n + uk−1n+1 − uk−1
n
h
−a (xn)uk+1
n − uk+1n−1 + uk−1
n − uk−1n−1
h
)= pk qn + f (tk, xn) ,qn = q (xn) , xn = nh,Nτ = T,
1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h
)+ ψ (xn) +
τ
2f (0, xn)
+τ
2
[1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ p0qn
],
1 6 n 6 M − 1,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = ζ (tk+1) ,−1 6 k 6 N − 1
(2.66)
Here, it is assumed that qM = q0 = 0, and∑M−1
i=1 qi , 0.We have the following theorem on
the stability of the difference scheme (2.66):
Theorem 2.3.7 For the solution of difference scheme (2.66), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1 + uh
k−12
N−1
k=1
Cτ(W2
2h)
53
6 M14 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M15 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.Now, we consider one more the second order of accuracy difference scheme
uk+1n − 2uk
n + uk−1n
τ2 −1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)+τ2
4
1h
a (xn+1)1h
[1h
(a (xn+2)
uk+1n+2 − uk+1
n+1h
− a (xn+1)uk+1
n+1 − uk+1n
h
)−
1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)]−
1h
a (xn)
[1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn−)
ukn − uk
n−1h
)−
1h
(a (xn)
uk+1n − uk+1
n−2h
− a (xn−1)uk+1
n−1 − uk+1n−2
h
)]= pk qn + f (tk, xn) ,qn = q (xn) , xn = nh,Nτ = T,
1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h
)+ ψ (xn) +
τ
2f (0, xn)
+τ
2
[1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ p0qn
],
1 6 n 6 M − 1,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = ζ (tk+1) ,−1 6 k 6 N − 1
(2.67)
Here, it is assumed that qM = q0 = 0, and∑M−1
i=1 qi , 0.We have the following theorem on
54
the stability of the difference scheme (2.45):
Theorem 2.3.8 For the solution of difference scheme (2.67), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk
N−1k=1
Cτ(W2
2h)
6 M14 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M15 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.
2.4 Numerical Experiments
In this section, we study the numerical solution of the identification problem
∂2u (t, x)∂t2 −
∂2u (t, x)∂x2 = p (t) sin x + e−t sin x,
x ∈ (0, π) , t ∈ (0,1) ,
u (0, x) = sin x,ut (0, x) = − sin x, x ∈ [0, π] ,
u (t,0) = u (t, π) = 0, t ∈ [0,1] ,∫ π
0 u (t, x) dx = 2e−t, t ∈ [0,1]
(2.68)
for a hyperbolic differential equation. The exact solution pair of this problem is (u (t, x) , p (t)) =(e−t sin x, e−t ) .Firstly, for the numerical solution of problem (2.68), we present the following first order of
accuracy difference scheme for the approximate solution for problem (2.68):
55
uk+1n − 2uk
n + uk−1n
τ2 −uk+1
n+1 − 2uk+1n + uk+1
n−1h2
= pk sin xn + e−tk+1 sin xn,
tk = kτ, xn = nh,1 6 k 6 N − 1,1 6 n 6 M − 1,
u0n = sin xn,
u1n − u0
n
τ= − sin xn,0 6 n 6 M,Mh = π,Nτ = 1,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = 2e−tk+1,−1 6 k 6 N − 1.
(2.69)
The algorithm for obtaining the solution of identification problem (2.69) contains three stages.
Actually, let us define
ukn = wk
n + ηk sin xn,0 ≤ k ≤ N,0 ≤ n ≤ M, (2.70)
Applying difference scheme (2.69) and formula (2.70), we will obtain
ηk+1 =2e−tk+1 −
∑M−1i=1 wk+1
i h∑M−1i=1 sin xih
,−1 ≤ k ≤ N − 1 (2.71)
and
wk+1n − 2wk
n + wk−1n
τ2 −wk+1
n+1 − 2wk+1n + wk+1
n−1h2
+
∑M−1i=1 wk+1
i h∑M−1i=1 sin xih
sin xn2(cos h − 1)
h2
=
[2∑M−1
i=1 sin xih
2(cos h − 1)h2 + 1
]e−tk+1 sin xn,
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = sin xn,
w1n − w
0n
τ= − sin xn,0 6 n 6 M,
wk+10 = wk+1
M = 0,−1 6 k 6 N − 1.
(2.72)
In the first stage, we find the solutionwk
nN
k=0
M
n=0of the corresponding first order of accuracy
difference scheme (2.72). For obtaining it, we will write difference scheme (2.72) in matrix
form as Awk+1 + Bwk + Cwk−1 = ϕk,1 6 k 6 N − 1,
w0 = sin xnMn=0 ,w
1 = (1 − τ) sin xnMn=0 ,
(2.73)
56
where A,B,C are (M + 1) × (M + 1) square matrices, ws, s = k, k ± 1, ϕk are (M + 1) × 1
column matrices, and
A =
1 0 0 · 0 0 0
b a + c1 b + c1 · c1 c1 0
0 b + c2 a + c2 · c2 c2 0
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
0 0 0 · 0 0 1
(M+1)×(M+1)
,
B =
0 0 0 · 0 0 0
0 e 0 · 0 0 0
0 0 e · 0 0 0
· · · · · · ·
0 0 0 · e 0 0
0 0 0 · 0 e 0
0 0 0 · 0 0 0
(M+1)×(M+1)
C =
0 0 0 · 0 0 0
0 g 0 · 0 0 0
0 0 g · 0 0 0
· · · · · · ·
0 0 0 · g 0 0
0 0 0 · 0 g 0
0 0 0 · 0 0 0
(M+1)×(M+1)
,
ϕk =
0
ϕk1
·
ϕkM−1
0
(M+1)×1
ws =
0
ws1
·
wsM−1
0
(M+1)×1
, for s = k, k ± 1.
Here,
a =1τ2 +
2h2 , b = −
1h2 , e = −
2τ2 ,g =
1τ2 ,
d =M−1∑i=1
sin xih, cn = sin xn2(cos h − 1)
dh,1 6 n 6 M − 1,
ϕkn =
[4(cos h − 1)
h2 ∑M−1i=1 sin xih
+ 1
]e−tk+1 sin xn,1 6 k 6 N − 1,1 6 n 6 M − 1.
57
So, we have the initial value problem for the second-order difference equation (2.73) with
respect to k with the matrix coefficients A,B and C. Since w0and w1are given, we can obtainwk
nN
k=0
M
n=0by (2.73).
Now, applying formula (2.31), we can obtain
pk =ηk+1 − 2ηk + ηk−1
τ2 ,1 ≤ k ≤ N − 1. (2.74)
In the second stage, we will obtain pkN−1k=1 by formulas (2.71) and (2.74). Finally, in the third
stage, we will obtain
uknN
k=0
M
n=0by formulas (2.70) and (2.71). The errors are computed by
Eu = max06k6N
(M−1∑n=1
u (tk, xn) − ukn
2 h
) 12
,
Ep = max16k6N−1
|p (tk) − pk | ,
where u (t, x) , p(t) represent the exact solution, ukn represents the numerical solutions at (tk, xn)
and pk represents the numerical solutions at tk . The numerical results are given in Table 1.
Table 1: Error analysis for difference scheme (2.69)
Error N =M = 20 N =M = 40 N =M = 80 N =M = 160
Eu 0.0347 0.0181 0.0092 0.0047
Ep 0.0462 0.0240 0.0123 0.0062
As can be seen in Table 1, if N and M are doubled, the value of errors between the exact
solution and approximate solution decreases by a factor of approximately 12 for the first order
difference scheme (2.69).
58
Secondly, for the numerical solution of problem (2.68), we present the following second order
of accuracy difference schemes for the approximate solution for problem (2.68):
uk+1n − 2uk
n + uk−1n
τ2 −12
ukn+1 − 2uk
n + ukn−1
h2
−14
[uk+1
n+1 − 2uk+1n + uk+1
n−1h2 +
uk−1n+1 − 2uk−1
n + uk−1n−1
h2
]= pk sin xn + e−tk sin xn,
tk = kτ,1 6 k 6 N − 1,Nτ = 1, xn = nh,1 6 n 6 M − 1,Mh = π,
u0n = sin xn,
u1n − u0
n
τ−τ
h2
(u1
n+1 − u0n+1 − 2u1
n + 2u0n + u1
n−1 − u0n−1
)= − sin xn +
τ
2
(u0
n+1 − 2u0n + u0
n−1h2 + p0 sin xn + sin xn
),0 6 n 6 M,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = 2e−tk+1,−1 6 k 6 N − 1
(2.75)
and
uk+1n − 2uk
n + uk−1n
τ2 −12
[uk+1
n+1 − 2uk+1n + uk+1
n−1h2 +
uk−1n+1 − 2uk−1
n + uk−1n−1
h2
]= pk sin xn + e−tk sin xn,
tk = kτ,1 6 k 6 N − 1,Nτ = 1, xn = nh,1 6 n 6 M − 1,Mh = π,
u0n = sin xn,
u1n − u0
n
τ−τ
h2
(u1
n+1 − u0n+1 − 2u1
n + 2u0n + u1
n−1 − u0n−1
)= − sin xn +
τ
2
(u0
n+1 − 2u0n + u0
n−1h2 + p0 sin xn + sin xn
),0 6 n 6 M,
uk+10 = uk+1
M = 0,∑M−1
i=1 uk+1i h = 2e−tk+1,−1 6 k 6 N − 1
(2.76)
The algorithm for obtaining the solution of identification problem (2.75) contains three stages.
Actually, let us define
ukn = wk
n + ηk sin xn,0 ≤ k ≤ N,0 ≤ n ≤ M, (2.77)
Applying the second order of accuracy difference scheme(2.75) and formula (2.77), we will
obtain
ηk =2e−tk −
∑M−1i=1 wk
i h∑M−1i=1 sin xih
,0 ≤ k ≤ N (2.78)
and the difference scheme
59
wk+1n − 2wk
n + wk−1n
τ2 −12wk
n+1 − 2wkn + w
kn−1
h2
−14
[wk+1
n+1 − 2wk+1n + wk+1
n−1h2 +
wk−1n+1 − 2wk−1
n + wk−1n−1
h2
]+
12(cos h − 1)
h2 ∑M−1i=1 sin xih
sin xn
×
(∑M−1i=1 wk+1
i h + 2∑M−1
i=1 wki h +
∑M−1i=1 wk−1
i h)
= e−tk sin xn +[e−tk+1 + 2e−tk + e−tk−1
] (cos h − 1)h2 ∑M−1
i=1 sin xihsin xn
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = sin xn,0 6 n 6 M,
w1n − w
0n
τ−τ
h2(w1
n+1 − 2w1n + w
1n−1
)+
2τ(cosh−1)h2 ∑M−1
i=1 sin xihsin xn
∑M−1i=1 w1
i h
+τ
2h2
(w0
n+1 − 2w0n + w
0n−1
)=
(τ2− 1
)sin xn
+4τe−t1(cosh−1)h2 ∑M−1
i=1 sin xihsin xn,0 6 n 6 M,
wk+10 = wk+1
M = 0,−1 6 k 6 N − 1.
(2.79)
In the first stage, we find the solutionwk
nN
k=0
M
n=0of the corresponding second order of
accuracy difference scheme (2.79). For obtaining it, we will write difference scheme (2.79),
in matrix form as Awk+1 + Bwk + Cwk−1 = ϕk,1 6 k 6 N − 1,
w0,w1 are given,(2.80)
where A,B,C are (M + 1) × (M + 1) square matrices, ws, s = k, k ± 1, ϕk are (M + 1) × 1
column matrices and
A =
1 0 0 · 0 0 0
b a + c1 b + c1 · c1 c1 c1
0 b + c2 a + c2 · c2 c2 c2
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
0 0 0 · 0 0 1
(M+1)×(M+1)
,
60
B =
0 0 0 · 0 0 0
r q + e1 r + e1 · e1 e1 e1
0 r + e2 q + e2 · e2 e2 e2
· · · · · · ·
0 eM−2 eM−2 · q + eM−2 r + eM−2 0
0 eM−1 eM−1 · r + eM−1 q + eM−1 r
0 0 0 · 0 0 0
(M+1)×(M+1)
,
C =
0 0 0 · 0 0 0
b a + c1 b + c1 · c1 c1 0
0 b + c2 a + c2 · c2 c2 0
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
0 0 0 · 0 0 0
(M+1)×(M+1)
,
ϕk =
0
ϕk1
·
ϕkM−1
0
(M+1)×1
, ws =
0
ws1
·
wsM−1
0
(M+1)×1
, for s = k, k ± 1,
w0 =
sin x0
sin x1
·
sin xM−1
sin xM
,
61
where,
a =1τ2 +
12h2 , b = −
14h2 ,r = −
12h2 ,q =
1h2 −
2τ2 ,
cn =12
sin xn(cos h − 1)
dh, en = sin xn
(cos h − 1)dh
,
d =M−1∑i=1
sin xih,1 6 n 6 M − 1,
ϕkn = e−tk sin xn +
(e−tk+1 + 2e−tk + e−tk−1
) (cos h − 1)h2 ∑M−1
i=1 sin xihsin xn,
1 6 k 6 N − 1,1 6 n 6 M − 1.
Finally, we obtain w1.Applying the formula
w1n − w
0n
τ−τ
h2(w1
n+1 − 2w1n + w
1n−1
)+
2τ(cos h − 1)h2 ∑M−1
i=1 sin xihsin xn
∑M−1i=1 w1
i h
+τ
2h2
(w0
n+1 − 2w0n + w
0n−1
)=
(τ2− 1
)sin xn +
4τe−t1(cos h − 1)h2 ∑M−1
i=1 sin xihsin xn,
and conditions w10 = w1
M = 0,we get
Ew1 + Fw0 = γ. (2.81)
Here
E =
1 0 0 · 0 0 0
y x + z1 y + z1 · z1 z1 0
0 y + z2 x + z2 · z2 z2 0
· · · · · · ·
0 zM−2 zM−2 · x + zM−2 y + zM−2 0
0 zM−1 zM−1 · y + zM−1 x + zM−1 y
0 0 0 · 0 0 1
(M+1)×(M+1)
,
62
x = −2τh2 +
1τ, y = −
τ
h2 , zn =2τ(cos h − 1)
dhsin xn,
u = −τ
2h2 , v =1τ+τ
h2 , v = −32τ−
2τh2 + 1,
F =
0 0 0 · 0 0 0
u v u · 0 0 0
0 u v · 0 0 0
· · · · · · ·
0 0 0 · v u 0
0 0 0 · u v u
0 0 0 · 0 0 0
(M+1)×(M+1)
,
γ =
0
∇1
·
∇M−1
0
(M+1)×1
,
∇n =
(τ
2− 1 +
4τe−t1(cos h − 1)h2 ∑M−1
i=1 sin xih
)sin xn.
From that, it follows
w1 = E−1(γ − Fw0
). (2.82)
So, we have the initial value problem for the second-order difference equation (2.80) with
respect to k with the matrix coefficients A,B and C. Since w0and w1are given, we can obtainwk
nN
k=0
M
n=0by (2.80)
Now, applying formula (2.49), we can obtain
pk =ηk+1 − 2ηk + ηk−1
τ2 ,1 ≤ k ≤ N − 1, p0 =2τ2η1 (2.83)
In the second stage, we will obtain pkN−1k=1 by formulas (2.78) and (2.83). Finally, in the third
stage, we will obtain
uknN
k=0
M
n=0by formulas (2.77) and (2.78).
63
In the same manner, we obtain algorithm for obtaining the solution of identification problem
(2.76) ukNk=0 =
uk
nN
k=0
M
n=0and pk
N−1k=1 by three stages. The errors are computed by
Eu = max06k6N
(M−1∑n=1
u (tk, xn) − ukn
2 h
) 12
,
Ep = max16k6N−1
|p (tk) − pk | ,
where u (t, x) , p(t) represent the exact solution, ukn represents the numerical solutions at
(tk, xn), and pk represents the numerical solutions at tk . The numerical results are given in the
following Tables 2 and 3.
Table 2:Error analysis for difference scheme (2.75)
Error N =M = 20 N =M = 40 N =M = 80 N =M = 160
Eu 0.0016 4.2029e − 04 1.0648e − 04 2.6796e − 05
Ep 0.0027 7.0457e − 04 1.7835e − 04 4.4867e − 05
Table 3: Error analysis for difference scheme (2.76)
Error N =M = 20 N =M = 40 N =M = 80 N =M = 160
Eu 0.0016 4.2011e − 04 1.0647e − 04 2.6795e − 05
Ep 0.0033 8.5650e − 04 2.1690e − 04 5.4570e − 05
As can be seen in Tables 2 and 3, if N and M are doubled, the value of errors between the exact
solution and approximate solution decreases by a factor of approximately 14 for the second
order difference schemes (2.75) and (2.76), respectively.
64
CHAPTER 3
STABILITY OF THE HYPERBOLIC DIFFERENTIAL AND DIFFERENCE
EQUATIONWITH NONLOCAL CONDITIONS
3.1 Introduction In this chapter, we consider the source identification problem for a
one-dimensional hyperbolic equation with nonlocal conditions
∂2u (t, x)∂t2 −
∂
∂x
(a (x)
∂u (t, x)∂x
)+ δu(t, x)
= p (t) q (x) + f (t, x) , x ∈ (0, l) , t ∈ (0,T) ,
u (0, x) = ϕ (x) ,ut (0, x) = ψ (x) , x ∈ [0, l] ,
u (t,0) = u (t, l) ,ux (t,0) = ux (t, l) ,∫ l0 u (t, x) dx = ζ (t) , t ∈ [0,T] ,
(3.1)
where u (t, x) and p (t) are unknown functions, a (x) ≥ a > 0,a (l) = a (0) , δ > 0,
f (t, x) , ζ (t) , ϕ (x) and ψ (x) are given sufficiently smooth functions and q (x) is a given
sufficiently smooth function assuming q (0) = q (l) , q′ (0) = q′ (l) and∫ l0 q (x) dx , 0.
3.2 Stability of the Differential Problem (3.1)
To formulate our results, we introduce the differential operator A defined by the formula
Au = −ddx
(a (x)
du (x)dx
)+ δu(x) (3.2)
with the domain
D (A) = u : u,u′′ ∈ L2 [0, l] ,u (0) = u (l) ,u′ (0) = u′ (l) .
It is easy that A is the self-adjoint positive-definite operator in H = L2 [0, l] . Actually, for all
u, v ∈ L2 [0, l] we have that
〈Au, v〉 =∫ l
0A(u)v (x) dx =
∫ l
0
[−
ddx
(a (x)
du (x)dx
)+ δu(x)
]v (x) dx
= −
(a (x)
du (x)dx
)v(x)
l0+
∫ l
0a (x)
du (x)dx
dv (x)dx
dx
+δ
∫ l
0u(x)v (x) dx
65
= −a (l)du (l)
dxv(l) + a (0)
du (0)dx
v(0) +∫ l
0a (x)
du (x)dx
dv (x)dx
dx
+δ
∫ l
0u(x)v (x) dx
Since
a (l) = a (0) ,u (0) = u (l) ,u′ (0) = u′ (l) , (3.3)
we have that
−a (l)du (l)
dxv(l) + a (0)
du (0)dx
v(0) = 0.
Therefore,
〈Au, v〉 =∫ l
0a (x)
du (x)dx
dv (x)dx
dx + δ∫ l
0u(x)v (x) dx.
Using (3.3), we get
〈u, Av〉 =∫ l
0u (x) A(v)dx =
∫ l
0u (x)
[−
ddx
(a (x)
dv (x)dx
)+ δv(x)
]dx
= − u(x)(a (x)
dv (x)dx
)l0+
∫ l
0a (x)
dv (x)dx
du (x)dx
dx
+δ
∫ l
0v(x)u (x) dx
= −u(l)a (l)dv (l)
dx+ u(0)a (0)
dv (0)dx
+
∫ l
0a (x)
dv (x)dx
du (x)dx
dx
=
∫ l
0a (x)
dv (x)dx
du (x)dx
dx + δ∫ l
0v(x)u (x) dx.
From that it follows
〈Au, v〉 = 〈u, Av〉
and
〈Au,u〉 =∫ l
0a (x)
du (x)dx
du (x)dx
dx + δ∫ l
0u(x)u (x) dx (3.4)
≥ δ
∫ l
0u(x)u (x) dx
= δ 〈u,u〉 .
We have the following theorem on the stability of problem (3.1).
66
Theorem 3.2.1 Assume that ϕ ∈ W22 [0, l] ,ψ ∈ W1
2 [0, l] and f (t, x) is a continuously
differentiable function in t and square integrable in x, and ζ (t) is a twice continuously
differentiable function. Suppose that q (x) is a sufficiently smooth function assuming q (0)
= q (l), q′ (0) = q′ (l) and∫ l0 q (x) dx , 0. Then, for the solution of problem (3.1), the
following stability estimates hold: ∂2u∂t2
C(L2[0,l])
+ ‖u‖C(W22 [0,l])
6 K1 (q)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l](3.5)
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
+ ‖ζ ′′‖C[0,T]
],
‖p‖C[0,T] 6 K2 (q)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l]+ ‖ζ ′′‖C[0,T] (3.6)
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
].
Proof. We will use the following substitution
u (t, x) = w (t, x) + η (t) q (x) , (3.7)
where η (t) is the function defined by formula
η (t) =∫ t
0(t − s) p (s) ds, η (0) = η′ (0) = 0. (3.8)
It is easy to see that w (t, x) is the solution of problem
∂2w (t, x)∂t2 −
∂
∂x
(a (x)
∂w (t, x)∂x
)+ δw(t, x)
= f (t, x) + η (t)[
ddx(a (x) q′ (x)) − δq (x)
],
x ∈ (0, l) , t ∈ (0,T) ,
w (0, x) = ϕ (x) ,wt (0, x) = ψ (x) , x ∈ [0, l] ,
w (t,0) = w (t, l) ,wx (t,0) = wx (t, l) , t ∈ [0,T] .
(3.9)
Applying the integral overdetermined condition∫ l0 u (t, x) dx = ζ (t) and substitution (3.7),
we get
η (t) =ζ (t) −
∫ l0 w (t, x) dx∫ l
0 q (x) dx.
67
From that and p (t) = η′′ (t) it follows
p (t) =ζ ′′ (t) −
∫ l0
∂2
∂t2w (t, x) dx∫ l0 q (x) dx
.
Applying∫ l0 q (x) dx , 0,we get estimate
|p (t)| ≤ K3 (q)
[|ζ ′′ (t)| +
∂2w (t, .)∂t2
L2[0,l]
](3.10)
for all t ∈ [0,T] . From that it follows
‖p‖C[0,T] 6 K3 (q)
[‖ζ ′′‖C[0,T] +
∂2w
∂t2
C(L2[0,l])
]. (3.11)
Now, using substitution (3.7), we get
∂2u (t, x)∂t2 =
∂2w (t, x)∂t2 + p (t) q (x) .
Applying the triangle inequality, we obtain ∂2u∂t2
C(L2[0,l])
6
∂2w
∂t2
C(L2[0,l])
+ ‖p‖C[0,T] ‖q‖L2[0,l] . (3.12)
Therefore, the proof of estimates (3.5) and (3.6) is based on equation (3.1), the triangle
inequality, estimates (3.11), (3.12) and on the following stability estimate ∂2w
∂t2
C(L2[0,l])
6 K4(q,a)[‖ϕ‖W2
2 [0,l]+ ‖ψ‖W1
2 [0,l](3.13)
+ ‖ f (0, .)‖L2[0,l] +
∂ f∂t
C(L2[0,l])
+ ‖ζ ′′‖C[0,T]
],
for the solution of problem (3.9). It was proved in Section 2.2 for the identification hyperbolic
problem with local boundary condition. The proof of (3.13) is carried out according to the
same approach. This completes the proof of Theorem 3.2.1.
3.3 Stability of the Difference Scheme
To formulate our results for the differential operator A defined by (3.2), we introduce the
difference operator Ah defined by the formula
Ahϕh (x) =
−
1h
(a (xn+1)
ϕn+1 − ϕn
h− a (xn)
ϕn − ϕn−1h
)+ δϕn
M−1
n=1,
68
acting in the space of grid functions ϕh (x) = ϕnMn=0 defined on [0, l]h satisfying the
conditions ϕ0 = ϕM , ϕ1 − ϕ0 = ϕM − ϕM−1.
It is easy that Ah is the self-adjoint positive-definite operator in H = L2h = L2 [0, l]h . Actually,
we have that ⟨Ahuh, vh⟩ = ∑
x∈[0,l]h
Ahuh(x)vh(x)h
=
M−1∑n=1
[−
1h
(a (xn+1)
un+1 − un
h− a (xn)
un − un−1h
)+ δun
]vnh
= −
M−1∑n=1
a (xn+1)un+1 − un
hvn +
M−1∑n=1
a (xn)un − un−1
hvn + δ
M−1∑n=1
unvnh
= −
M∑n=2
a (xn)un − un−1
hvn−1 +
M−1∑n=1
a (xn)un − un−1
hvn + δ
M−1∑n=1
unvnh
= −a (xM)uM − uM−1
hvM−1 + a (x1)
u1 − u0h
v0
−
M−1∑n=2
a (xn)un − un−1
hvn−1 +
M−1∑n=2
a (xn)un − un−1
hvn + δ
M−1∑n=1
unvnh
= −a (xM)uM − uM−1
hvM + a (x1)
u1 − u0h
v0
+
M∑n=2
a (xn)un − un−1
hvn − vn−1
hh + δ
M−1∑n=1
unvnh.
Since
a (xM) = a (x1) , ϕ0 = ϕM, ϕ1 − ϕ0 = ϕM − ϕM−1, (3.14)
we have that
−a (xM)uM − uM−1
hvM + a (x1)
u1 − u0h
v0 = 0.
⟨Ahuh, vh⟩ = M∑
n=2a (xn)
un − un−1h
vn − vn−1h
h + δM−1∑n=1
unvnh.
Using (3.14), we get ⟨uh, Ahv
h⟩ = ∑x∈[0,l]h
uh(x)Ahvh(x)h
69
=
M−1∑n=1
un
[−
1h
(a (xn+1)
vn+1 − vn
h− a (xn)
vn − vn−1h
)+ δvn
]h
= −
M−1∑n=1
una (xn+1)vn+1 − vn
h+
M−1∑n=1
una (xn)vn − vn−1
h+ δ
M−1∑n=1
unvnh
= −
M∑n=2
un−1a (xn)vn − vn−1
h+
M−1∑n=1
una (xn)vn − vn−1
h+ δ
M−1∑n=1
unvnh
= −uM−1a (xM)vM − vM−1
h+ u1a (x1)
v1 − v0h−
M−1∑n=2
a (xn)vn − vn−1
hun−1
+
M−1∑n=2
a (xn) unvn − vn−1
hh + δ
M−1∑n=1
unvnh
= a (xM)uM − uM−1
hvM − vM−1
h+
M−1∑n=2
a (xn)un − un−1
hvn − vn−1
hh
+δ
M−1∑n=1
unvnh
=
M∑n=2
a (xn)un − un−1
hvn − vn−1
hh + δ
M−1∑n=1
unvnh.
From that it follows ⟨Ahuh, vh⟩ = ⟨
uh, Ahvh⟩
and ⟨Ahuh,uh⟩ = M∑
n=2a (xn)
un − un−1h
un − un−1h
h + δM−1∑n=1
ununh
≥ δ
M−1∑n=1
ununh
= δ⟨uh,uh⟩ .
3.3.1 The first order of accuracy difference scheme
For the numerical solution
uknN
k=0
M
n=0of problem (3.1), we consider the first order of
accuracy difference scheme
70
uk+1n − 2uk
n + uk−1n
τ2 −
(a (xn+1)
uk+1n+1 − uk+1
n
h2 − a (xn)uk+1
n − uk+1n−1
h2
)+δuk+1
n = pk q (xn) + f (tk, xn) , tk = kτ, xn = nh,
1 6 k 6 N − 1,1 6 n 6 M − 1,Nτ = T,
u0n = ϕ (xn) ,
u1n − u0
n
τ= ψ (xn) ,0 6 n 6 M,Mh = l,
uk+10 = uk+1
M ,uk+11 − uk+1
0 = uk+1M − uk+1
M−1,∑M−1i=1 uk+1
i h = ζ (tk+1) ,−1 6 k 6 N − 1.
(3.15)
Here, it is assumed that
q0 = qM,q1 − q0 = qM − qM−1
and∑M−1
i=1 qi , 0.We have the following theorem on the stability of difference scheme (3.15).
Theorem 3.3.1 For the solution of difference scheme (3.15)
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1
N−1k=1
Cτ(W2
2h)(3.16)
6 M5 (q)[ ϕh
W2
2h+
ψh
W12h+
f h1
L2h
+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M6 (q)[ ϕh
W2
2h+
ψh
W22h+
f h1
L2h(3.17)
+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.Here and throughout this subsection f h
k (x) = f (tk, xn)Mn=0 ,1 ≤ k ≤ N − 1.
Proof. We will use the following substitution
ukn = wk
n + ηk qn, (3.18)
where
qn = q (xn) ,
71
and
ηk+1 =
k∑i=1(k + 1 − i) piτ
2,1 6 k 6 N − 1, η0 = η1 = 0. (3.19)
It is easy to see thatwk
nN
k=0
M
n=0is the solution of difference problem
wk+1n − 2wk
n + wk−1n
τ2 −1h
(a (xn+1)
wk+1n+1 − w
k+1n
h− a (xn)
wk+1n − wk+1
n−1h
)+δwk+1
n = f (tk, xn) +1h
[a (xn+1)
qn+1 − qn
h− a (xn)
qn − qn−1h
− δhqn
]ηk+1,
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = ϕ (xn) ,
w1n − w
0n
τ= ψ (xn) ,0 6 n 6 M,
wk+10 = wk+1
M ,wk+11 − wk+1
0 = wk+1M − wk+1
M−1,−1 6 k 6 N − 1.
(3.20)
Applying the overdetermined condition∑M−1
i=1 uk+1i h = ζ (tk+1) and substitution ( 3.18), one
can obtain that
ηk+1 =ζk+1 −
∑M−1i=1 wk+1
i h∑M−1i=1 qih
. (3.21)
Then, using formulas pk =ηk+1 − 2ηk + ηk−1
τ2 and (3.21), we get
pk =ζk+1 − 2ζk + ζk−1 −
∑M−1i=1
(wk+1
i − 2wki + w
k−1i
)h
τ2 ∑M−1i=1 qih
.
|pk | 6 K7 (q)
[ ζk+1 − 2ζk + ζk−1
τ2
+ whk+1 − 2wh
k + whk−1
τ2
L2h
](3.22)
for all 1 6 k 6 N − 1. From that it follows
pkN−1k=1
C[0,T]τ
≤ K7 (q) ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T ]τ
+
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
. (3.23)
Now, using substitution (3.18), we get
uk+1n − 2uk
n + uk−1n
τ2 =wk+1
n − 2wkn + w
k−1n
τ2 + pk q (xn) .
72
Applying the triangle inequality, we obtain
uk+1i − 2uk
i + uk−1i
τ2
N−1
k=1
M
n=0
Cτ(L2h)
(3.24)
6
wk+1
i − 2wki + w
k−1i
τ2
N−1
k=1
M
n=0
Cτ(L2h)
+ pk
N−1k=1
C[0,T]τ
q (xn)M−1n=1
L2h
.
Therefore, the proof of estimates (3.16) and (3.17) is based on equation (3.15), the triangle
inequality, estimates (3.23), (3.24) and on the following stability estimate wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
(3.25)
6 K8(q)[ ϕh
W2
2h+
ψh
W12h+
f h1
L2h
+
f hk − f h
k−1τ
N−1
k=2
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
for the solution of difference problem (3.20). It was proved in Section 2.3 for the identification
hyperbolic problemwith local boundary condition. The proof of (3.25) is carried out according
to the same approach. This completes the proof of Theorem 3.3.1.
3.3.2 The second order of accuracy difference scheme
We have the following approximate formulas
−u(2h) + 4u(h) − 3u(0)2h
− ux(0) = o(h2),
3u(l) − 4u(l − h) + u(l − 2h)2h
− ux(l) = o(h2).
(3.26)
Applying (3.26) for the numerical solution
uknN
k=0
M
n=0of problem (3.1), we consider the
second order of accuracy difference scheme
73
uk+1n − 2uk
n + uk−1n
τ2 −12h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)−
14h
(a (xn+1)
uk+1n+1 − uk+1
n + uk−1n+1 − uk−1
n
h
−a (xn)uk+1
n − uk+1n−1 + uk−1
n − uk−1n−1
h
)+
12δuk
n + δuk+1
n + uk−1n
4
= pk qn + f (tk, xn) , xn = nh, tk = kτ,
Nτ = T,1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h+ δh(u0
n − u1n)
)= ψ (xn) +
τ
2[ f (0, xn) + p0qn]
−τ
2
[−
1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ δu0
n
],1 6 n 6 M − 1,
uk+10 = uk+1
M ,−uk+12 + 4uk+1
1 − 3uk+10 = 3uk+1
M − 4uk+1M−1 + uk+1
M−2,∑M−1i=1 uk+1
i h = ζ (tk+1) ,−1 6 k 6 N − 1
(3.27)
Here, it is assumed that
qM = q0,−q2 + 4q1 − 3q0 = 3qM − 4qM−1 + qM−2 (3.28)
and∑M−1
i=1 qi , 0.We have the following theorem on the stability of the difference scheme
(3.27):
Theorem 3.3.2 For the solution of difference scheme (3.27), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1 + 2uh
k + uhk−1
4
N−1
k=1
Cτ(W2
2h)
(3.29)
6 M9 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M10 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h(3.30)
74
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.Here and throughout this subsection f h
k (x) = f (tk, xn)Mn=0 ,1 ≤ k ≤ N − 1.
Proof. We will use the substitution
ukn = wk
n + ηk qn, (3.31)
where
ηk =
k∑i=1
(k − i) pi + (k − (i − 1)) pi−1
2
τ2,1 6 k 6 N, η0 = 0, (3.32)
It is easy to see thatwk
nN
k=0
M
n=0is the solution of the difference problem
wk+1n − 2wk
n + wk−1n
τ2 −12h
(a (xn+1)
wkn+1 − w
kn
h− a (xn)
wkn − w
kn−1
h
)−
14h
(a (xn+1)
wk+1n+1 − w
k+1n + wk−1
n+1 − wk−1n
h
−a (xn)wk+1
n − wk+1n−1 + w
k−1n − wk−1
n−1h
)+
12δwk
n + δwk+1
n + wk−1n
4
=1h
[a (xn+1)
qn+1 − qn
h− a (xn)
qn − qn−1h
− δhqn
] 14(ηk+1 + 2ηk + ηk−1)
+ f (tk, xn) ,1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = ϕ (xn) ,0 6 n 6 M,
w1n − w
0n
τ−τ
h
(a (xn+1)
w1n+1 − w
0n+1 − w
1n + w
0n
h
−a (xn)w1
n − w0n − w
1n−1 + w
0n−1
h+ δh(w0
n − w1n)
)−τ
h
(a (xn+1)
qn+1 − qn
h
−a (xn)qn − qn−1
h− δhqn
)η1 = ψ (xn) +
τ
2f (0, xn)
−τ
2
[−
1h
(a (xn+1)
w0n+1 − w
0n
h− a (xn)
w0n − w
0n−1
h
)+ δw0
n
],1 6 n 6 M − 1,
wk+10 = wk+1
M ,−wk+12 + 4wk+1
1 − 3wk+10 = 3wk+1
M − 4wk+1M−1 + w
k+1M−2,
−1 6 k 6 N − 1.
(3.33)
Now, we will take an estimate for |pk | .Using the overdetermined condition∑M−1
i=1 uk+1i h =
75
ζ (tk+1) and substitution (3.31), one can obtain
ηk =ζk −
∑M−1i=1 wk
i h∑M−1i=1 qih
. (3.34)
Then, using the formulas pk =ηk+1−2ηk+ηk−1
τ2 and (3.34), we get
pk =ζk+1 − 2ζk + ζk−1 −
∑M−1i=1
(wk+1
i − 2wki + w
k−1i
)h
τ2 ∑M−1i=1 qih
.
Then, applying the discrete analogue of the Cauchy–Schwarz inequality and the triangle
inequality, we obtain
|pk | 6 M11 (q)
[ ζk+1 − 2ζk + ζk−1
τ2
+ whk+1 − 2wh
k + whk−1
τ2
L2h
](3.35)
for all 1 6 k 6 N − 1. From that, it follows
pkN−1k=1
C[0,T]τ
6 K11 (q) ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
(3.36)
+
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
.Now, using substitution (3.31), we get
uk+1n − 2uk
n + uk−1n
τ2 =wk+1
n − 2wkn + w
k−1n
τ2 + pk q (xn) .
Applying the triangle inequality, we obtain
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
6
wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
(3.37)
+ pk
N−1k=1
C[0,T]τ
q (xn)Mn=0
L2h
.
Therefore, the proof of estimates (3.29) and (3.30) is based on equation (3.27), the triangle
inequality, estimates (3.36), (3.37) and on the following stability estimate wh
k+1 − 2whk + w
hk−1
τ2
N−1
k=1
Cτ(L2h)
6 M12(q)[ ϕh
W2
2h+
ψh
W12h
(3.38)
76
+ f h
0
L2h+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.
.for the solution of difference problem (3.33). It was proved in Section 2.3 for the identification
hyperbolic problemwith local boundary condition. The proof of (3.38) is carried out according
to the same approach. This completes the proof of Theorem 3.3.2.
Applying ( 3.26) and same approach with this work we can study another the second order of
accuracy difference scheme
uk+1n − 2uk
n + uk−1n
τ2 −12h
(a (xn+1)
uk+1n+1 − uk+1
n + uk−1n+1 − uk−1
n
h
−a (xn)uk+1
n − uk+1n−1 + uk−1
n − uk−1n−1
h
)+ δ
uk+1n + uk−1
n
2
= pk qn + f (tk, xn) ,qn = q (xn) , xn = nh,Nτ = T,
1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h+ δh(u0
n − u1n)
)= ψ (xn) +
τ
2[ f (0, xn) + p0qn]
−τ
2
[−
1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ δu0
n
],1 6 n 6 M − 1,
uk+10 = uk+1
M ,−uk+12 + 4uk+1
1 − 3uk+10 = 3uk+1
M − 4uk+1M−1 + uk+1
M−2,∑M−1i=1 uk+1
i h = ζ (tk+1) ,−1 6 k 6 N − 1
(3.39)
Here, it is assumed that the formula (3.28) holds and∑M−1
i=1 qi , 0.We have the following
theorem on the stability of the difference scheme (3.39):
Theorem 3.3.3 For the solution of difference scheme (3.39), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk+1 + uh
k−12
N−1
k=1
Cτ(W2
2h)
6 M14 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
77
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M15 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.
Now, we consider one more the second order of accuracy difference scheme
uk+1n − 2uk
n + uk−1n
τ2 −1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)+ δuk
n
+τ2
4
1h
a (xn+1)1h
[1h
(a (xn+2)
uk+1n+2 − uk+1
n+1h
− a (xn+1)uk+1
n+1 − uk+1n
h
)−
1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn)
ukn − uk
n−1h
)]−
1h
a (xn)
[1h
(a (xn+1)
ukn+1 − uk
n
h− a (xn−)
ukn − uk
n−1h
)−
1h
(a (xn)
uk+1n − uk+1
n−2h
− a (xn−1)uk+1
n−1 − uk+1n−2
h
)]−2δ
1h
(a (xn+1)
uk+1n+1 − uk+1
n
h− a (xn)
uk+1n − uk+1
n−1h
)+ δ2uk+1
n
= pk qn + f (tk, xn) ,qn = q (xn) , xn = nh,Nτ = T,
1 6 n 6 M − 1,Mh = l, tk = kτ,1 6 k 6 N − 1,
u0n = ϕ (xn) ,0 6 n 6 M,
u1n − u0
n
τ−τ
h
(a (xn+1)
u1n+1 − u0
n+1 − u1n + u0
n
h
−a (xn)u1
n − u0n − u1
n−1 + u0n−1
h+ δh(u0
n − u1n)
)= ψ (xn) +
τ
2[ f (0, xn) + p0qn]
−τ
2
[−
1h
(a (xn+1)
u0n+1 − u0
n
h− a (xn)
u0n − u0
n−1h
)+ δu0
n
],1 6 n 6 M − 1,
uk+10 = uk+1
M ,−uk+12 + 4uk+1
1 − 3uk+10 = 3uk+1
M − 4uk+1M−1 + uk+1
M−2,∑M−1i=1 uk+1
i h = ζ (tk+1) ,−1 6 k 6 N − 1
(3.40)
78
Here, it is assumed that formula (3.28) holds and∑M−1
i=1 qi , 0.We have the following theorem
on the stability of the difference scheme (3.40):
Theorem 3.3.4 For the solution of difference scheme (3.40), the following stability estimates
hold:
uhk+1 − 2uh
k + uhk−1
τ2
N−1
k=1
Cτ(L2h)
+
uhk
N−1k=1
Cτ(W2
2h)
6 M14 (q)[ ϕh
W2
2h+
ψh
W12h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
, pkN−1k=1
C[0,T]τ
6 M15 (q)[ ϕh
W2
2h+
ψh
W22h+
f h0
L2h
+
f hk − f h
k−1τ
N−1
k=1
Cτ(L2h)
+
ζk+1 − 2ζk + ζk−1
τ2
N−1
k=1
C[0,T]τ
.
3.4 Numerical Experiments
In this section, we study the numerical solution of the identification problem
∂2u (t, x)∂t2 −
∂2u (t, x)∂x2 = p (t) (4 + 4 sin 2x) + 4e−2t sin 2x,
x ∈ (0, π) , t ∈ (0,1) ,
u (0, x) = 1 + sin 2x,ut (0, x) = −2 (1 + sin 2x) , x ∈ [0, π] ,
u (t,0) = u (t, π) ,ux (t,0) = ux (t, π) t ∈ [0,1] ,∫ π
0 u (t, x) dx = πe−2t, t ∈ [0,1]
(3.41)
for a hyperbolic differential equation with non-local conditions. The exact solution pair of this
problem is (u (t, x) , p (t)) =(e−2t (1 + sin 2x) , e−2t ) .
Firstly, for the numerical solution of problem (3.41), we present the following first order of
accuracy difference scheme for the approximate solution for problem (3.41):
79
uk+1n − 2uk
n + uk−1n
τ2 −uk+1
n+1 − 2uk+1n + uk+1
n−1h2
= pk (4 + 4 sin 2xn) + 4e−2tk+1 sin 2xn,
tk = kτ, xn = nh,1 6 k 6 N − 1,1 6 n 6 M − 1,
u0n = 1 + sin 2xn,
u1n − u0
n
τ= −2 (1 + sin 2xn) ,
0 6 n 6 M,Mh = π,Nτ = 1,
uk+10 = uk+1
M ,uk+11 − uk+1
0 = uk+1M − uk+1
M−1,∑M−1i=1 uk+1
i h = πe−2tk+1,−1 6 k 6 N − 1.
(3.42)
Algorithm for obtaining the solution of identification problem (3.42) contains three stages.
Actually, let us define
ukn = wk
n + 4ηk (1 + sin 2xn) ,0 ≤ k ≤ N,0 ≤ n ≤ M, (3.43)
Applying difference scheme (3.42) and formula (3.43), we will obtain formula
ηk+1 =πe−2tk+1 −
∑M−1i=1 wk+1
i h
4∑M−1
i=1 (1 + sin 2xi) h,−1 ≤ k ≤ N − 1 (3.44)
and the difference scheme
wk+1n − 2wk
n + wk−1n
τ2 −wk+1
n+1 − 2wk+1n + wk+1
n−1h2
+
∑M−1i=1 wk+1
i h∑M−1i=1 (1 + sin 2xi) h
sin 2xn2(cos 2h − 1)
h2
=
[2π(cos 2h − 1)
h2 ∑M−1i=1 (1 + sin 2xi) h
+ 4
]e−2tk+1 sin 2xn,
1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = 1 + sin 2xn,
w1n − w
0n
τ= −2 (1 + sin 2xn) ,0 6 n 6 M,
wk+10 = wk+1
M ,wk+11 − wk+1
0 = wk+1M − wk+1
M−1,
−1 6 k 6 N − 1.
(3.45)
In the first stage, we find numerical solutionwk
nN
k=0
M
n=0of corresponding first order of
accuracy auxiliary difference scheme (3.45). For obtaining the solution of difference scheme
(3.45), we will write it in the matrix form as
80
Awk+1 + Bwk + Cwk−1 = ϕk,1 6 k 6 N − 1,
w0 = 1 + sin 2xnMn=0 ,w
1 = (1 − 2τ) 1 + sin 2xnMn=0,
(3.46)
where A,B,C are (M + 1) × (M + 1) square matrices, ws, s = k, k ± 1, f k are (M + 1) × 1
column matrices and
A =
1 0 0 · 0 0 −1
b a + c1 b + c1 · c1 c1 0
0 b + c2 a + c2 · c2 c2 0
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
−1 1 0 · 0 1 −1
(M+1)×(M+1)
,
B =
0 0 0 · 0 0 0
0 e 0 · 0 0 0
0 0 e · 0 0 0
· · · · · · ·
0 0 0 · e 0 0
0 0 0 · 0 e 0
0 0 0 · 0 0 0
(M+1)×(M+1)
C =
0 0 0 · 0 0 0
0 g 0 · 0 0 0
0 0 g · 0 0 0
· · · · · · ·
0 0 0 · g 0 0
0 0 0 · 0 g 0
0 0 0 · 0 0 0
(M+1)×(M+1)
,
ϕk =
0
ϕk1
·
ϕkM−1
0
(M+1)×1
ws =
0
ws1
·
wsM−1
0
(M+1)×1
, for s = k, k ± 1.
Here,
a =1τ2 +
2h2 , b = −
1h2 , cn =
2(cos 2h − 1)hd
sin 2xn,
d =M−1∑i=1(1 + sin 2xi) h, e = −
2τ2 ,g =
1τ2 ,
81
ϕkn =
[2π(cos 2h − 1)
h3 ∑M−1i=1 (1 + sin 2xi)
+ 4
]e−2tk+1 sin 2xn,1 6 k 6 N − 1,1 6 n 6 M − 1.
So, we have the initial value problem for the second order difference equation (3.46) with
respect to k with matrix coefficients A,B and C. Since w0and w1are given, we can obtainwk
nN
k=0
M
n=0by (3.46).
Now, applying formula (3.32), we can obtain
pk =ηk+1 − 2ηk + ηk−1
τ2 ,1 ≤ k ≤ N − 1. (3.47)
In the second stage, we will obtain pkN−1k=1 by formulas (3.44) and (3.47). Finally, in the third
stage, we will obtain
uknN
k=0
M
n=0by formulas (3.43) and (3.44). The errors are computed by
Eu = max06k6N
(M−1∑n=1
u (tk, xn) − ukn
2 h
) 12
,
Ep = max16k6N−1
|p (tk) − pk | ,
where u (t, x) , p(t) represent the exact solution, ukn represent the numerical solutions at (tk, xn)
and pk represent the numerical solutions at tk . The numerical results are given in the following
table.
Table 4: Error analysis for difference scheme (3.42)
Error N =M = 20 N =M = 40 N =M = 80 N =M = 160
Eu 0.0560 0.0289 0.0147 0.0075
Ep 0.0476 0.0244 0.0123 0.0062
As can be seen in Table 4, if N and M are doubled, the value of errors between the exact
solution and approximate solution decreases by a factor of approximately 1/2 for the first
order difference scheme (3.42).
82
Secondly, for the numerical solution of problem (3.41), we present the following second order
of accuracy difference scheme for the approximate solution for problem (3.41):
uk+1n − 2uk
n + uk−1n
τ2 −12
ukn+1 − 2uk
n + ukn−1
h2
−14
[uk+1
n+1 − 2uk+1n + uk+1
n−1h2 +
uk−1n+1 − 2uk−1
n + uk−1n−1
h2
]= pk (4 + 4 sin 2xn) + 4e−2tk+1 sin 2xn,
tk = kτ, xn = nh,1 6 k 6 N − 1, ,1 6 n 6 M − 1,
u0n = 1 + sin 2xn,
u1n − u0
n
τ−τ
h2
(u1
n+1 − u0n+1 − 2u1
n + 2u0n + u1
n−1 − u0n−1
)= −2 (1 + sin 2xn) +
τ
2
u0
n+1 − 2u0n + u0
n−1h2
+4p0 (1 + sin 2xn) + 4 sin 2xn 0 6 n 6 M,Mh = π,Nτ = 1,
uk+10 = uk+1
M ,
−uk+12 + 4uk+1
1 − 3uk+10 = 3uk+1
M − 4uk+1M−1 + uk+1
M−2,∑M−1i=1 uk+1
i h = πe−tk+1,−1 6 k 6 N − 1
(3.48)
The algorithm for obtaining the solution of identification problem (3.48) contains three stages.
Actually, let us define
ukn = wk
n + 4ηk (1 + sin 2xn) ,0 ≤ k ≤ N,0 ≤ n ≤ M, (3.49)
Applying the second order of accuracy difference scheme(2.75) and formula (2.77), we will
obtain
ηk =πe−2tk −
∑M−1i=1 wk
i h
4∑M−1
i=1 (1 + sin 2xi) h,0 ≤ k ≤ N (3.50)
and the difference scheme
83
wk+1n − 2wk
n + wk−1n
τ2 −12wk
n+1 − 2wkn + w
kn−1
h2
−14
[wk+1
n+1 − 2wk+1n + wk+1
n−1h2 +
wk−1n+1 − 2wk−1
n + wk−1n−1
h2
]+
12
(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn
×
(∑M−1i=1 wk+1
i h + 2∑M−1
i=1 wki h +
∑M−1i=1 wk−1
i h)
= 4e−2tk sin xn +[e−2tk+1 + 2e−2tk + e−2tk−1
]×
12
π(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn,1 6 k 6 N − 1,1 6 n 6 M − 1,
w0n = 1 + sin 2xn,0 6 n 6 M,
w1n − w
0n
τ−τ
h2(w1
n+1 − 2w1n + w
1n−1
)+
2τ(cosh−1)h2 ∑M−1
i=1 sin xihsin 2xn
∑M−1i=1 w1
i h
+τ
2h2
(w0
n+1 − 2w0n + w
0n−1
)= 2τ sin 2xn − 2 (1 + sin 2xn)
+2τπe−2t1(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn,0 6 n 6 M,
wk+10 = wk+1
M ,
−wk+12 + 4wk+1
1 − 3wk+10 = 3wk+1
M − 4wk+1M−1 + w
k+1M−2,
−1 6 k 6 N − 1.
(3.51)
In the first stage, we find the solutionwk
nN
k=0
M
n=0of the corresponding second-order-of-
accuracy difference scheme (3.51). For obtaining it, we will write difference scheme (3.51),
in matrix form as Awk+1 + Bwk + Cwk−1 = ϕk =,1 6 k 6 N − 1,
w0,w1 are given,(3.52)
where A,B,C are (M + 1) × (M + 1) square matrices, ws, s = k, k ± 1, ϕk are (M + 1) × 1
column matrices and
84
A =
1 0 0 · 0 0 0
b a + c1 b + c1 · c1 c1 c1
0 b + c2 a + c2 · c2 c2 c2
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
0 0 0 · 0 0 1
(M+1)×(M+1)
,
B =
0 0 0 · 0 0 0
r q + e1 r + e1 · e1 e1 e1
0 r + e2 q + e2 · e2 e2 e2
· · · · · · ·
0 eM−2 eM−2 · q + eM−2 r + eM−2 0
0 eM−1 eM−1 · r + eM−1 q + eM−1 r
0 0 0 · 0 0 0
(M+1)×(M+1)
,
C =
0 0 0 · 0 0 0
b a + c1 b + c1 · c1 c1 0
0 b + c2 a + c2 · c2 c2 0
· · · · · · ·
0 cM−2 cM−2 · a + cM−2 b + cM−2 0
0 cM−1 cM−1 · b + cM−1 a + cM−1 b
0 0 0 · 0 0 0
(M+1)×(M+1)
,
ϕk =
0
ϕk1
·
ϕkM−1
0
(M+1)×1
, ws =
0
ws1
·
wsM−1
0
(M+1)×1
, for s = k, k ± 1.
85
Here,
a =1τ2 +
12h2 , b = −
14h2 , ,r = −
12h2 ,q =
1h2 −
2τ2 ,
cn =12
sin xn(cos 2h − 1)
dh, en = sin xn
(cos 2h − 1)dh
,
d =M−1∑i=1(1 + sin xi) h,1 6 n 6 M − 1,
ϕkn = 4e−2tk sin xn +
(e−2tk+1 + 2e−2tk + e−2tk−1
)(cos 2h − 1)
h2 ∑M−1i=1 (1 + sin xi) h
sin 2xn,
1 6 k 6 N − 1,1 6 n 6 M − 1.
Finally, we obtain w1.Applying the formula
w1n − w
0n
τ−τ
h2(w1
n+1 − 2w1n + w
1n−1
)+
2τ(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn
∑M−1i=1 w1
i h
+τ
2h2
(w0
n+1 − 2w0n + w
0n−1
)= 2τ sin 2xn − 2 (1 + sin 2xn) +
2τπe−2t1(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn,
and conditions w10 = w1
M,w11 − w
10 = w1
M − w1M−1we get
Ew1 + Fw0 = γ. (3.53)
Here
E =
1 0 0 · 0 0 0
y x + z1 y + z1 · z1 z1 0
0 y + z2 x + z2 · z2 z2 0
· · · · · · ·
0 zM−2 zM−2 · x + zM−2 y + zM−2 0
0 zM−1 zM−1 · y + zM−1 x + zM−1 y
0 0 0 · 0 0 1
(M+1)×(M+1)
,
x = −2τh2 +
1τ, y = −
τ
h2 , zn =2τ(cos 2h − 1)
dhsin 2xn,
u = −τ
2h2 , v =1τ+τ
h2 , v = −32τ−
2τh2 + 1,
86
F =
0 0 0 · 0 0 0
u v u · 0 0 0
0 u v · 0 0 0
· · · · · · ·
0 0 0 · v u 0
0 0 0 · u v u
0 0 0 · 0 0 0
(M+1)×(M+1)
,
γ =
0
∇1
·
∇M−1
0
(M+1)×1
,
∇n = 2τ sin 2xn − 2 (1 + sin 2xn) +2τπe−2t1(cos 2h − 1)h2 ∑M−1
i=1 (1 + sin xi) hsin 2xn.
From that, it follows
w1 = E−1(γ − Fw0
). (3.54)
So, we have the initial value problem for the second-order difference equation (3.52) with
respect to k with the matrix coefficients A,B and C. Since w0and w1are given, we can obtainwk
nN
k=0
M
n=0by (3.52). Now, applying formula (3.32), we can obtain
pk =ηk+1 − 2ηk + ηk−1
τ2 ,1 ≤ k ≤ N − 1, p0 =2τ2η1 (3.55)
In the second stage, we will obtain pkN−1k=1 by formulas (3.50) and (3.55). Finally, in the third
stage, we will obtain
uknN
k=0
M
n=0by formulas (3.49) and (3.50).
The errors are computed by
Eu = max06k6N
(M−1∑n=1
u (tk, xn) − ukn
2 h
) 12
,
Ep = max16k6N−1
|p (tk) − pk | ,
87
where u (t, x) , p(t) represent the exact solution, ukn represents the numerical solutions at
(tk, xn), and pk represents the numerical solutions at tk . The numerical results are given in the
following Table.
Table 5: Error analysis for difference scheme (3.48)
Error N =M = 20 N =M = 40 N =M = 80 N =M = 160
Eu 0.00254 6.355e − 04 1.5675e − 04 3.9723e − 05
Ep 0.0028 7.0523e − 04 1.76857e − 04 4.4758e − 05
As can be seen in Table 5, if N and M are doubled, the value of errors between the exact
solution and approximate solution decreases by a factor of approximately 14 for the second-order
difference scheme (3.48).
88
CHAPTER 4
CONCLUSION
This thesis is devoted to study the source identification problem for hyperbolic differential
equations with unknown parameter p(t). The following results are obtained:
• The history of direct and inverse boundary value problems for hyperbolic differential
equations are studied.
• Fourier series, Laplace transform and Fourier transform methods are applied for the
solution of several identification problems for hyperbolic differential equations.
• The theorem on the stability estimates for the solution of the source identification
problem for hyperbolic differential equations with local and nonlocal conditions is proved.
• The first and second order of accuracy difference schemes for the approximate solution
of the one dimensional identification problem for hyperbolic differential equation with local
and nonlocal conditions are presented.
• The theorem on the stability estimates for these difference schemes for the numerical
solution of identification problems for hyperbolic differential equations with local and nonlocal
conditions is established.
• The Matlab implementation of these difference schemes is presented.
• The theoretical statements for the solution of these difference schemes are supported
by the results of numerical examples.
89
90
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95
Appendix 1
Matlab Programming
Matlab Implementation of Difference Schemes (2.69)
clc; clear all ; close all;
N=160;
M=160;
h=pi/M; tau=1/N;
a=(1/(tau^2))+(2/(h^2));
e=-2/(tau^2);
b=-1/(h^2);
g=1/(tau^2);
d=0;
for i=1:M-1;
d=d+h*sin(i*h);
end;
z=2*(cos(h)-1)/(d*h);
A=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
A(i,j)=z*sin((i-1)*h);
end;
end;
for i=2:M
A(i,i)=a+(z*sin((i-1)*h));
end;
for i=2:M-1;
A(i,i+1)=b+(z*sin((i-1)*h));
end;
for i=3:M;
96
A(i,i-1)=b+(z*sin((i-1)*h));
end;
A(1,1)=1;A(M+1,M+1)=1;A(2,1)=b;A(M,M+1)=b;
A;
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=e;
end;
B;
C=zeros(M+1,M+1);
for n=2:M;
C(n,n)=g;
end;
C;
fii=zeros(M+1,1);
for j=1:M+1;
for k=2:N;
fii(j,k)=((4*(cos(h)-1)/(d*(h^2)))+1)*exp(-k*tau)*sin((j-1)*h);
end;
end;
fii;
G=inv(A);
W=zeros(M+1,1);
for j=1:M+1;
W(j,1)=sin((j-1)*h);
W(j,2)=(1-tau)*sin((j-1)*h);
for k=3:N+1;
W(:,k)=G*(-(B*W(:,k-1))-(C*W(:,k-2))+fii(:,k-1));
end;
end;
for k=2:N;
97
D=0;
for j=1:M-1;
S(j)=D+W(j,k+1)-2*(W(j,k))+W(j,k-1);
D=S(j);
end;
p(k)=(2*exp(-(k+1)*tau)-4*exp(-k*tau)+2*exp(-(k-1)*tau)-(h*D))/(d*(tau^2));
end;
p(k);
L=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
L(i,j)=0;
end
end;
for i=2:M;
L(i,i)=a;
end;
for i=2:M-1;
L(i,i+1)=b;
end;
for i=3:M;
L(i,i-1)=b;
end;
L(1,1)=1;
L(M+1,M+1)=1;
L;
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=e;
end;
B;
98
C=zeros(M+1,M+1);
for n=2:M;
C(n,n)=g;
end;
C;
fii=zeros(M+1,1) ;
for j=1:M+1;
for k=2:N;
x=(j-1)*h;
fii(j,k)=exp(-k*tau)*sin(x)+(p(k)*sin(x));
end;
end;
fii;
G=inv(L);
u=zeros(M+1,1);
for j=1:M+1;
x=(j-1)*h;
u(j,1)=sin(x);
u(j,2)=(1-tau)*sin(x);
end;
for k=3:N+1;
u(:,k)=G*(-(B*u(:,k-1))-(C*u(:,k-2))+fii(:,k-1));
end;
%\%\%\%\%\%'EXACT SOLUTION OF THIS PDE' \%\%\%\%\%\%\%\%
for j=1:M+1;
for k=1:N+1;
t=(k-1)*tau;
x=(j-1)*h;
es(j,k)=(1-t)*sin(x);
eu(j,k)=exp(-t)*sin(x);
end;
99
end;
for k=2:N;
t=(k-1)*tau;
ep(k)=exp(-t);
end;
% ABSOLUTE DIFFERENCES ;
absdiff=max(max(abs(es-W)))
absdiff=max(max(abs(ep-p)))
absdiff=max(max(abs(eu-u)))
100
Appendix 2
Matlab Programming
Matlab Implementation of Difference Schemes (2.75)
clc; clear all ; close all;
N=20;M=20;
h=pi/M; tau=1/N;
d=0;
for j=1:M-1;
d=d+h*sin((j)*h);
end;
a=1/(tau^2)+1/(2*(h^2));q=1/(h^2)-2/(tau^2);
b=-1/(4*(h^2));r=-1/(2*(h^2));
c=(cos(h)-1)/(2*d*h);e=(cos(h)-1)/(d*h);
A=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
A(i,j)=c*sin((i-1)*h);
end;
end;
for i=2:M
A(i,i)=a+c*sin((i-1)*h);
end;
for i=2:M-1;
A(i,i+1)=b+c*sin((i-1)*h);
end;
for i=3:M;
A(i,i-1)=b+c*sin((i-1)*h);
end;
A(1,1)=1;A(M+1,M+1)=1;A(2,1)=b;A(M,M+1)=b;
101
A;
B=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
B(i,j)=e*sin((i-1)*h);
end;
end;
for i=2:M
B(i,i)=q+e*sin((i-1)*h);
end;
for i=2:M-1;
B(i,i+1)=r+e*sin((i-1)*h);
end;
for i=3:M;
B(i,i-1)=r+e*sin((i-1)*h);
end;
B(2,1)=r;B(M,M+1)=r;
B;
C=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
C(i,j)=c*sin((i-1)*h);
end;
end;
for i=2:M
C(i,i)=a+c*sin((i-1)*h);
end;
for i=2:M-1;
C(i,i+1)=b+c*sin((i-1)*h);
end;
for i=3:M;
102
C(i,i-1)=b+c*sin((i-1)*h);
end;
C(2,1)=b;C(M,M+1)=b;
C;
fii=zeros(M+1,1);
for j=1:M+1;
for k=2:N;
fii(j,k)=exp(-(k-1)*tau)*sin((j-1)*h)+(exp(-k*tau)+2*exp(-(k-1)*tau)+exp(-(k-
2)*tau))*sin((j-1)*h)*((cos(h)-1)/(d*(h^2)));
end;
end;
e1=-tau/(h^2);e2=(2*tau)/(h^2)+1/tau;
y=(2*tau*(cos(h)-1))/(d*h);
E=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
E(i,j)=y*sin((i-1)*h);
end;
end;
for i=2:M
E(i,i)=e2+y*sin((i-1)*h);
end;
for i=2:M-1;
E(i,i+1)=e1+y*sin((i-1)*h);
end;
for i=3:M;
E(i,i-1)=e1+y*sin((i-1)*h);
end;
E(1,1)=1;E(M+1,M+1)=1;E(2,1)=e1;E(M,M+1)=e1;
E;
f1=tau/(2*(h^2));f2=-tau/(h^2)-1/tau;
103
F=zeros(M+1,M+1);
for i=2:M
F(i,i)=f2;
end;
for i=2:M-1;
F(i,i+1)=f1;
end;
for i=3:M;
F(i,i-1)=f1;
end;
F(2,1)=f1;F(M,M+1)=f1;
F;
phy=zeros(M+1,1);
for j=1:M+1;
phy(j)=(tau/2-1)*sin((j-1)*h)+((2*y)/h)*(exp(-tau)*sin((j-1)*h));
end;
Z=inv(E);G=inv(A);
W=zeros(M+1,1);
for j=1:M+1;
W(j,1)=sin((j-1)*h);
end;
W(:,2)=Z*(phy-F*W(:,1));
for k=3:N+1;
W(:,k)=G*(-(B*W(:,k-1))-(C*W(:,k-2))+fii(:,k-1));
end;
for k=2:N;
D=0;
for j=1:M-1;
S(j)=D+W(j,k+1)-2*(W(j,k))+W(j,k-1);
D=S(j);
104
end;
p(k)=(2*exp(-(k)*tau)-(4*exp(-(k-1)*tau))+(2*exp(-(k-2)*tau))-(h*D))/(d*(tau^2));
end;
p(k);
R=zeros(M+1,M+1);
for i=2:M
R(i,i)=a;
end;
for i=2:M-1;
R(i,i+1)=b;
end;
for i=3:M;
R(i,i-1)=b;
end;
R(1,1)=1;R(M+1,M+1)=1;R(2,1)=b;R(M,M+1)=b;
R;
L=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
L(i,j)=0;
end;
end;
for i=2:M
L(i,i)=q;
end;
for i=2:M;
L(i,i+1)=r;
end;
for i=3:M;
L(i,i-1)=r;
end;
105
L(2,1)=r;L(M,M+1);
L;
Q=zeros(M+1,M+1);
for i=2:M
Q(i,i)=a;
end;
for i=2:M-1;
Q(i,i+1)=b;
end;
for i=3:M;
Q(i,i-1)=b;
end;
Q(2,1)=b;Q(M,M+1)=b;
Q;
fii=zeros(M+1,1) ;
for j=2:M;
for k=2:N;
x=(j-1)*h;
fii(j,k)=exp(-(k-1)*tau)*sin(x)+(p(k)*sin(x));
end;
end;
G=inv(R);
u=zeros(M+1,1);
for k=3:N+1;
for j=1:M+1;
x=(j-1)*h;
u(j,1)=sin(x);
u(j,2)=(1-tau+((tau^2)/(2)))*sin((j-1)*h);
u(:,k)=G*(-(L*u(:,k-1))-(Q*u(:,k-2))+fii(:,k-1));
end;
end;
106
%\%\%\%\%\%'EXACT SOLUTION OF THIS PDE' \%\%\%\%\%\%\%\%
for j=1:M+1;
for k=1:N+1;
t=(k-1)*tau; x=(j-1)*h;
es(j,k)=(1-t)*sin(x);
eu(j,k)=exp(-t)*sin(x);
end;
end;
for k=2:N;
t=(k-1)*tau;
ep(k)=exp(-t);
end
% ABSOLUTE DIFFERENCES ;
absdiff=max(max(abs(es-W)))
absdiff=max(max(abs(ep-p)))
absdiff=max(max(abs(eu-u)))
107
Appendix 3
Matlab Programming
Matlab Implementation of Difference Schemes (2.76)
clc; clear all ; close all;
N=160;
M=160;
h=pi/M; tau=1/N;
d=0;
for i=1:M-1;
d=d+h*sin(i*h);
end;
a=(1/(tau^2))+(1/(h^2));b=-1/(2*(h^2));g=-2/(tau^2);c=(cos(h)-1)/(d*h);
A=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
A(i,j)=c*sin((i-1)*h);
end;
end;
for i=2:M
A(i,i)=a+(c*sin((i-1)*h));
end;
for i=2:M-1;
A(i,i+1)=b+(c*sin((i-1)*h));
end;
for i=3:M;
A(i,i-1)=b+(c*sin((i-1)*h));
end;
A(1,1)=1;
A(M+1,M+1)=1;
108
A(2,1)=b;
A(M,M+1)=b;
A;
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=g;
end;
B;
C=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
C(i,j)=c*sin((i-1)*h);
end;
end;
for i=2:M
C(i,i)=a+(c*sin((i-1)*h));
end;
for i=2:M-1;
C(i,i+1)=b+(c*sin((i-1)*h));
end;
for i=3:M;
C(i,i-1)=b+(c*sin((i-1)*h));
end;
C(2,1)=b;
C(M,M+1)=b;
C;
fii=zeros(M+1,1);
for j=1:M+1;
for k=2:N;
fii(j,k)=(exp(-(k-1)*tau)+((2*c)/h)*exp(-k*tau)+((2*c)/h)*exp(-(k-2)*tau))*sin((j-1)*h);
109
end;
end;
e1=-tau/(h^2);
e2=(2*tau)/(h^2)+1/tau;
y=(2*tau*(cos(h)-1))/(d*h);
E=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
E(i,j)=y*sin((i-1)*h);
end;
end;
for i=2:M
E(i,i)=e2+y*sin((i-1)*h);
end;
for i=2:M-1;
E(i,i+1)=e1+y*sin((i-1)*h);
end;
for i=3:M;
E(i,i-1)=e1+y*sin((i-1)*h);
end;
E(1,1)=1;
E(M+1,M+1)=1;
E(2,1)=e1;
E(M,M+1)=e1;
E;
f1=tau/(2*(h^2));
f2=-tau/(h^2)-1/tau;
F=zeros(M+1,M+1);
for i=2:M
F(i,i)=f2;
end;
110
for i=2:M-1;
F(i,i+1)=f1;
end;
for i=3:M;
F(i,i-1)=f1;
end;
F(2,1)=f1;
F(M,M+1)=f1;
F;
phy=zeros(M+1,1);
for j=1:M+1;
phy(j)=(tau/2-1)*sin((j-1)*h)+((2*y)/h)*(exp(-tau)*sin((j-1)*h));
end;
Z=inv(E);
G=inv(A);
W=zeros(M+1,1);
for j=1:M+1;
W(j,1)=sin((j-1)*h);
end;
W(:,2)=Z*(phy-F*W(:,1));
for k=3:N+1;
W(:,k)=G*(-(B*W(:,k-1))-(C*W(:,k-2))+fii(:,k-1));
end;
for k=2:N;
D=0;
for j=1:M-1;
S(j)=D+W(j,k+1)-2*(W(j,k))+W(j,k-1);
D=S(j);
end;
p(k)=(2*exp(-(k)*tau)-(4*exp(-(k-1)*tau))+(2*exp(-(k-2)*tau))-(h*D))/(d*(tau^2));
end;
111
p(k);
R=zeros(M+1,M+1);
for i=2:M
R(i,i)=a;
end;
for i=2:M-1;
R(i,i+1)=b;
end;
for i=3:M;
R(i,i-1)=b;
end;
R(1,1)=1;
R(M+1,M+1)=1;
R(2,1)=b;
R(M,M+1)=b;
R;
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=g;
end;
B;
Q=zeros(M+1,M+1);
for i=2:M
Q(i,i)=a;
end;
for i=2:M-1;
Q(i,i+1)=b;
end;
for i=3:M;
Q(i,i-1)=b;
end;
112
Q(2,1)=b;
Q(M,M+1)=b;
Q ;
fii=zeros(M+1,1) ;
for j=2:M;
for k=2:N;
x=(j-1)*h;
fii(j,k)=exp(-(k-1)*tau)*sin(x)+(p(k)*sin(x));
end;
end;
G=inv(R);
u=zeros(M+1,1);
for k=3:N+1;
for j=1:M+1;
x=(j-1)*h;
u(j,1)=sin(x);
u(j,2)=(1-tau+((tau^2)/(2)))*sin((j-1)*h);
u(:,k)=G*(-(B*u(:,k-1))-(Q*u(:,k-2))+fii(:,k-1));
end;
end;
%\%\%\%\%\%'EXACT SOLUTION OF THIS PDE' \%\%\%\%\%\%\%\%
for j=1:M+1;
for k=1:N+1;
t=(k-1)*tau;
x=(j-1)*h;
es(j,k)=(1-t)*sin(x);
% ep(k)=exp(-t);
eu(j,k)=exp(-t)*sin(x);
end;
end;
for k=2:N;
113
t=(k-1)*tau;
ep(k)=exp(-t);
end
% ABSOLUTE DIFFERENCES ;
absdiff=max(max(abs(es-W)))
absdiff=max(max(abs(ep-p)))
absdiff=max(max(abs(eu-u)))
114
Appendix 4
Matlab Programming
Matlab Implementation of Difference Schemes (3.42)
N=160;M=160;
h=pi/M; tau=1/N;
a=(1/(tau^2))+(2/(h^2));e=-2/(tau^2);b=-1/(h^2);g=1/(tau^2);
r=0;
for i=1:M-1;
r=r+h*(1+sin(2*(i)*h));
end;
r;
z=2*(cos(2*h)-1)/(r*(h));
A=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
A(i,j)=z*sin(2*(i-1)*h);
end;
end;
for i=2:M;
A(i,i)=a+(z*sin(2*(i-1)*h));
end;
for i=2:M-1;
A(i,i+1)=b+(z*sin(2*(i-1)*h));
end;
for i=3:M;
A(i,i-1)=b+(z*sin(2*(i-1)*h));
end;
A(1,1)=1; A(M+1,M+1)=-1; A(1,M+1)=-1; A(M+1,1)=-1; A(M+1,M)=1; A(M+1,2)=1;
A(2,1)=b; A(M,M+1)=b;A;
115
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=e;
end;
B;
C=zeros(M+1,M+1);
for n=2:M;
C(n,n)=g;
end;
C;
fii=zeros(M+1,1) ;
for j=2:M;
for k=2:N;
t=(k)*tau; x=(j-1)*h;
fii(j,k)=((2*(pi)*(cos(2*h)-1)/(r*(h^2)))+4)*exp(-2*(k)*tau)*sin(2*(j-1)*h);
end;
end;
fii;
G=inv(A);
W=zeros(M+1,1);
forj=1:M+1;
x=(j-1)*h;
W(j,1)=1+sin(2*(j-1)*h);
W(j,2)=(1-2*tau)*(1+sin(2*(j-1)*h));
for k=3:N+1;
W(:,k)=G*(-(B*W(:,k-1))-(C*W(:,k-2))+fii(:,k-1));
end;
end;
D=0;
for k=2:N;
for j=1:M-1;
116
S(j)=D+(W(j,k+1)-2*(W(j,k))+W(j,k-1));
D=S(j);
end;
p(k)=(pi)*(exp(-2*(k-1)*tau))*(1/r)-((h*D)/(4*r)*(tau^2));
end;
p(k);
L=zeros(M+1,M+1);
for i=2:M;
L(i,i)=a;
end;
for i=2:M;
L(i,i+1)=b;
end;
for i=2:M;
L(i,i-1)=b;
end;
L(1,1)=1; L(M+1,M+1)=-1; L(1,M+1)=-1;L(M+1,1)=-1; L(M+1,M)=1; L(M+1,2)=1;
L;
B=zeros(M+1,M+1);
for n=2:M;
B(n,n)=e;
end;
B;
C=zeros(M+1,M+1);
for n=2:M;
C(n,n)=g;
end;
C;
fii=zeros(M+1,M+1) ;
for j=2:M;
for k=2:N;
117
x=(j-1)*h; t=(k)*tau;
fii(j,k)=4*exp(-2*t)*sin(2*(j-1)*h)+p(k)*(4+4*sin(2*(j-1)*h));
end;
end;
fii;
G=inv(L);
u=zeros(M+1,1);
for j=1:M+1;
x=(j-1)*h;
u(j,1)=(1+sin(2*(j-1)*h));
u(j,2)=(1-2*tau)*(1+sin(2*(j-1)*h));
for k=3:N+1;
u(:,k)=G*(-(B*u(:,k-1))-(C*u(:,k-2))+fii(:,k-1));
end;
end;
%\%\%\%\%\%'EXACT SOLUTION OF THIS PDE' \%\%\%\%\%\%\%\%
for j=1:M+1;
for k=1:N+1;
t=(k-1)*tau; x=(j-1)*h;
es(j,k)=(-2*t+1)*(1+sin(2*(j-1)*h));
eu(j,k)=exp(-2*t)*(1+sin(2*(j-1)*h));
end;
end;
for k=2:N;
t=(k-1)*tau;
ep(k)=exp(-2*t);
end;
% ABSOLUTE DIFFERENCES ;
absdiff=max(max(abs(es-W)))
absdiff=max(max(abs(ep-p)))
absdiff=max(max(abs(eu-u)))
118
Appendix 5
Matlab Programming
Matlab Implementation of Difference Schemes (3.48)
clc; clear all ; close all;
N=80;M=80;
h=pi/M; tau=1/N;
d=0;
for i=1:M;
d=d+h*(1+sin(2*(i-1)*h));
end;
a=(1/(tau^2))+(1/(2*(h^2)));
b=-1/(4*(h^2));
q=(1/(h^2))-(2/(tau^2));
r=-1/(2*(h^2));
e=(cos(2*h)-1)/(d*h);
c=(cos(2*h)-1)/(2*d*h);
A=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
A(i,j)=c*sin(2*(i-1)*h);
end;
end;
for i=2:M
A(i,i)=a+c*sin(2*(i-1)*h);
end;
for i=2:M-1;
A(i,i+1)=b+c*sin(2*(i-1)*h);
end;
for i=3:M;
119
A(i,i-1)=b+c*sin(2*(i-1)*h);
end;
A(1,1)=1; A(1,M+1)=-1; A(2,1)=b; A(M,M+1)=b;
A(M+1,1)=-3; A(M+1,2)=4; A(M+1,3)=-1;
A(M+1,M-1)=-1;A(M+1,M)=4;A(M+1,M+1)=-3;
B=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
B(i,j)=e*sin(2*(i-1)*h);
end;
end;
for i=2:M
B(i,i)=q+(e*sin(2*(i-1)*h));
end;
for i=2:M-1;
B(i,i+1)=r+(e*sin(2*(i-1)*h));
end;
for i=3:M;
B(i,i-1)=r+(e*sin(2*(i-1)*h));
end;
B(2,1)=r;B(M,M+1)=r;
B;
C=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
C(i,j)=c*sin(2*(i-1)*h);
end;
end;
for i=2:M
C(i,i)=a+(c*sin(2*(i-1)*h));
end;
120
for i=2:M-1;
C(i,i+1)=b+(c*sin(2*(i-1)*h));
end;
for i=3:M;
C(i,i-1)=b+(c*sin(2*(i-1)*h));
end;
C(2,1)=b;C(M,M+1)=b;
fii=zeros(M+1,1);
for j=2:M;
for k=1:N+1;
fii(j,k)=(4*exp(-2*(k-1)*tau))*sin(2*(j-1)*h)
+((c*(pi))/h)*(exp(-2*(k)*tau)+2*exp(-2*(k-1)*tau)+exp(-2*(k-2)*tau))*sin(2*(j-1)*h);
end;
end;
ee1=-tau/(h^2);
ee2=(2*tau)/(h^2)+1/tau;
y=((2*tau)*(cos(2*h)-1))/(d*h);
E=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
E(i,j)=y*sin(2*(i-1)*h);
end;
end;
for i=2:M
E(i,i)=ee2+y*sin(2*(i-1)*h);
end;
for i=2:M-1;
E(i,i+1)=ee1+y*sin(2*(i-1)*h);
end;
for i=3:M;
E(i,i-1)=ee1+y*sin(2*(i-1)*h);
121
end;
E(1,1)=1; E(1,M+1)=-1; E(2,1)=ee1; E(M,M+1)=ee2;
E(M+1,1)=-3; E(M+1,2)=4;E(M+1,3)=-1;
E(M+1,M-1)=-1;E(M+1,M)=4;E(M+1,M+1)=-3;
f1=tau/(2*(h^2));f2=-tau/(h^2)-1/tau;
F=zeros(M+1,M+1);
for i=2:M
F(i,i)=f2;
end;
for i=2:M-1;
F(i,i+1)=f1;
end;
for i=3:M;
F(i,i-1)=f1;
end;
F(2,1)=f1;
F(M,M+1)=f1
phy=zeros(M+1,1);
for j=1:M+1;
phy(j)=-2*(1+sin(2*(j-1)*h))+(2*tau)*sin(2*(j-1)*h)+((y*pi)/h)*(exp(-2*tau))*sin(2*(j-
1)*h);
end;
Z=inv(E);
G=inv(A);
W=zeros(M+1,1);
for j=1:M+1;
W(j,1)=1+sin(2*(j-1)*h);
end;
W(:,2)=Z*(phy-F*W(:,1));
for k=3:N+1;
W(:,k)=G*(-(B*W(:,k-1))-(C*W(:,k-2))+fii(:,k-1));
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end;
for k=2:N;
D=0;
for j=1:M-1;
S(j)=D+W(j,k+1)-2*(W(j,k))+W(j,k-1);
D=S(j);
end;
p(k)=((pi)*(exp(-(k)*tau)-2*exp(-(k-1)*tau)+exp(-(k-2)*tau))-(h*D))/(4*d*(tau^2));
end;
p(k);
R=zeros(M+1,M+1);
for i=2:M
R(i,i)=a;
end;
for i=2:M-1;
R(i,i+1)=b;
end;
for i=3:M;
R(i,i-1)=b;
end;
R(1,1)=1;
R(M+1,M+1)=1;
R(2,1)=b;
R(M,M+1)=b;
R;
L=zeros(M+1,M+1);
for i=2:M;
for j=2:M;
L(i,j)=0;
end;
end;
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for i=2:M
L(i,i)=q;
end;
for i=2:M;
L(i,i+1)=r;
end;
for i=3:M;
L(i,i-1)=r;
end;
L(2,1)=r;
L(M,M+1);
L;
Q=zeros(M+1,M+1);
for i=2:M
Q(i,i)=a;
end;
for i=2:M-1;
Q(i,i+1)=b;
end;
for i=3:M;
Q(i,i-1)=b;
end;
Q(2,1)=b;
Q(M,M+1)=b;
Q ;
fii=zeros(M+1,M+1) ;
for j=2:M;
for k=2:N;
x=(j-1)*h;
t=(k)*tau;
fii(j,k)=4*exp(-2*t)*sin(2*(j-1)*h)+4*p(k)*(1+sin(2*(j-1)*h));
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end;
end;
fii;
G=inv(R);
u=zeros(M+1,1);
for k=3:N+1;
for j=1:M+1;
x=(j-1)*h;
u(j,1)=1+sin(2*(j-1)*h);
u(j,2)=(1-2*tau+2*(tau^2))*(1+sin(2*(j-1)*h));
u(:,k)=G*(-(L*u(:,k-1))-(Q*u(:,k-2))+fii(:,k-1));
end;
end;
%\%\%\%\%\%'EXACT SOLUTION OF THIS PDE' \%\%\%\%\%\%\%\%
for j=1:M+1;
for k=1:N+1;
t=(k-1)*tau;
x=(j-1)*h;
es(j,k)=(-2*t+1)*(1+sin(2*(j-1)*h));
eu(j,k)=exp(-2*t)*(1+sin(2*(j-1)*h));
end;
end;
for k=2:N;
t=(k-1)*tau;
ep(k)=exp(-2*t);
end;
% ABSOLUTE DIFFERENCES ;
absdiff=max(max(abs(es-W)))
absdiff=max(max(abs(ep-p)))
absdiff=max(max(abs(eu-u)))